Equations involving derivatives - AP Calculus AB

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Question

Practicing the chain rule level 1 E!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

$y(x)'=\frac{4}{4x-3}$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

.

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

, $g(x)=\sqrt{sin(x)}$ and .

The reason why we did not define $g(x)=\sqrt{x}$ , and

is because of the property mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}ln(g(h(x)))$

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{2}ln(sin(x^2))$ since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}cos(x^2)*2x$ which is

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

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Question

Practicing the chain rule level 3 A!

Find the derivative of the function

$y(x)=ln(\sqrt{sin(x^2)})$

Tap to reveal answer

Answer

To understand why the answer is

,

first remember that .

Then, you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

.

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

, $g(x)=\sqrt{sin(x)}$ and .

The reason why we did not define $g(x)=\sqrt{x}$ , and

is because of the property mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}ln(g(h(x)))$

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{2}ln(sin(x^2))$ since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}cos(x^2)*2x$ which is

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

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Question

Practicing the chain rule level 3 B!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

$q(z)=cos(cos(tan(z)))sin(tan(z))sec^{2}(z)$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

$q'(z)=sec^{2}(z)$ .

can be treated as a composition of the functions

, and .

in terms of , and is actually which means

since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(z) first, then differentiate it

Step 2: Look at the next function g(z), keep it inside the other function f'(z)

Step 3: Look at the next function h(z), keep it inside the other function g(z)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(z) but multiply it by the factors f'(g(h(z)))g'(h(z))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(z), g(z), g'(z), h(z) and h'(z) for the expressions you found before:

$cos(cos(tan(z)))sin(tan(z)))*sec^{2}(z)$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to differentiate. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

← Didn't Know|Knew It →

Question

Practicing the chain rule level 3 C!

Find the derivative of the function

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$

Tap to reveal answer

Answer

To understand why the answer is

$y'(x)=\frac{4x}{(x^2+3)ln^2(\frac{1}{(x^2+3)^2})}$ ,

Then, you must understand that the derivative of

$\large y(x)=\frac{1}{x}$

is actually

$\large y'(x)=\frac{-1}{x^2}$ .

Next, you must understand that the derivative of

$\large y(x)=ln(x)$

is actually

$\large y'(x)=\frac{1}{x}$ .

And finally, you must understand that the derivative of

$\frac{1}{(2x+3)^2}$

is actually

$y'(x)=\frac{-4x}{(x^2+3)^3}$ .

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x+a}$ , and .

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{ln((x^2+3)^{-2})}$ since in $\frac{1}{x+a}$ is substituted with and in can be substituted with $\frac{1}{(x^2+3)^2}$ .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\large \large y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

← Didn't Know|Knew It →

Question

Find the eqation of the line tangent to the graph of $y=x+x^{1/2}$ at point .

Tap to reveal answer

Answer

$y=x+x^{1/2}$

$y'=1+\frac{1}{2}x^{-1/2}$

$y'=1+\frac{1}{2(x^{1/2})}$

Now we plug in .

$y'=1+\frac{1}{2(4^{1/2})} = 1+\frac{1}{4} = \frac{5}{4}$ this is the slope.

Now use the point slope formula to find the tangent line.

$y-6=\frac{5}{4}(x-4)$

$y-6=\frac{5}{4}x-5$

$y=\frac{5}{4}x+1$

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Question

Practicing the chain rule level 1 A!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

can be treated as a composition of the functions

and .

in terms of and is actually , which means

since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 1 B!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

can be treated as a composition of the functions

and .

in terms of and is actually , which means

since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the component composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 1 C!

Find the derivative of the function

$y(t)=e^{2x}$

Tap to reveal answer

Answer

To understand why the answer is

$y'(x)=2e^{2x}$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

$y(x)=e^{2x}$ can be treated as a composition of the functions

and .

in terms of and is actually , which means

$f(g(x))=e^{2x}$ since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the component composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$y'(x)=2e^{2x}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 1 D!

Find the derivative of the function

$p(x)=\frac{2}{2-3x}$

Tap to reveal answer

Answer

To understand why the answer is

$p'(x)=\frac{6}{(2-3x)^2}$ ,

you must understand that the derivative of

$y(x)=\frac{1}{x}$

is actually

$y(x)'=\frac{-1}{x^2}$ .

Next, you must understand that the derivative of

is actually

.

$p(x)=\frac{2}{-3x+2}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x}$ and .

in terms of and is actually and

$f(g(x))=\frac{2}{-3x+2}$ since in $\frac{1}{x}$ is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the component composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the function g(x) but multiply its derivative by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$p'(x)=\frac{6}{(2-3x)^2}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 2 A!

Find the derivative of the function

$p(x)=e^{^{x^{3}}}$

Tap to reveal answer

Answer

To understand why the answer is

$p'(x)=3x^{2}e^{x^3}$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

$p(x)=e^{x^{3}}$ can be treated as a composition of the functions

and $g(x)=x^{3}$ .

in terms of and is actually and

$f(g(x))=e^{x^{3}}$ since in is substituted with $x^{3}$ .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the function g(x) but multiply its derivative by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$p'(x)=3x^{2}e^{x^{3}}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 2 B!

Find the derivative of the function

$p(t)=tan^{2}(t)$

Tap to reveal answer

Answer

To understand why the answer is

,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

can be treated as a composition of the functions

and .

in terms of and is actually which means

since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(t), g(t) and g'(t) for the expressions you found before:

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 2 C!

Find the derivative of the function

$y(x)=\frac{-1}{2^{x}-4}$

Tap to reveal answer

Answer

To understand why the answer is

$y'(x)=\frac{2^xln(2)}{(2^x-4)^2}$

you must understand that the derivative of

$y(x)=\frac{-1}{x}$

is actually

$y'(x)=\frac{1}{x^2}$ .

Next, you must understand that the derivative of

is actually

.

$y(x)=\frac{-1}{2^x-4}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x}$ and .

in terms of and is actually which means

$f(g(x))=\frac{-1}{2^x-4}$ since in $\frac{-1}{x}$ is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$y'(x)=\frac{2^xln(2)}{(2^x-4)^2}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 2 D!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

$y'(x)=\frac{-1}{\sqrt{1-x^2}arccos(x)}$ ,

you must understand that the derivative of

is actually

$y'(x)=\frac{1}{x}$ .

Next, you must understand that the derivative of

is actually

$y'(x)=\frac{-1}{\sqrt{1-x^2}}$ .

can be treated as a composition of the functions

$f(x)=\frac{1}{x}$ and .

in terms of and is actually which means

since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$y'(x)=\frac{-1}{\sqrt{1-x^2}arccos(x)}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

← Didn't Know|Knew It →

Question

Practicing the chain rule level 1 E!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

$y(x)'=\frac{4}{4x-3}$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

.

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

, $g(x)=\sqrt{sin(x)}$ and .

The reason why we did not define $g(x)=\sqrt{x}$ , and

is because of the property mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}ln(g(h(x)))$

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{2}ln(sin(x^2))$ since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}cos(x^2)*2x$ which is

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

← Didn't Know|Knew It →

Question

Practicing the chain rule level 3 A!

Find the derivative of the function

$y(x)=ln(\sqrt{sin(x^2)})$

Tap to reveal answer

Answer

To understand why the answer is

,

first remember that .

Then, you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

.

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

, $g(x)=\sqrt{sin(x)}$ and .

The reason why we did not define $g(x)=\sqrt{x}$ , and

is because of the property mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}ln(g(h(x)))$

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{2}ln(sin(x^2))$ since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}cos(x^2)*2x$ which is

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

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Question

Practicing the chain rule level 3 B!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

$q(z)=cos(cos(tan(z)))sin(tan(z))sec^{2}(z)$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

$q'(z)=sec^{2}(z)$ .

can be treated as a composition of the functions

, and .

in terms of , and is actually which means

since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(z) first, then differentiate it

Step 2: Look at the next function g(z), keep it inside the other function f'(z)

Step 3: Look at the next function h(z), keep it inside the other function g(z)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(z) but multiply it by the factors f'(g(h(z)))g'(h(z))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(z), g(z), g'(z), h(z) and h'(z) for the expressions you found before:

$cos(cos(tan(z)))sin(tan(z)))*sec^{2}(z)$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to differentiate. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

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Question

Practicing the chain rule level 3 C!

Find the derivative of the function

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$

Tap to reveal answer

Answer

To understand why the answer is

$y'(x)=\frac{4x}{(x^2+3)ln^2(\frac{1}{(x^2+3)^2})}$ ,

Then, you must understand that the derivative of

$\large y(x)=\frac{1}{x}$

is actually

$\large y'(x)=\frac{-1}{x^2}$ .

Next, you must understand that the derivative of

$\large y(x)=ln(x)$

is actually

$\large y'(x)=\frac{1}{x}$ .

And finally, you must understand that the derivative of

$\frac{1}{(2x+3)^2}$

is actually

$y'(x)=\frac{-4x}{(x^2+3)^3}$ .

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x+a}$ , and .

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{ln((x^2+3)^{-2})}$ since in $\frac{1}{x+a}$ is substituted with and in can be substituted with $\frac{1}{(x^2+3)^2}$ .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\large \large y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

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Question

Find the eqation of the line tangent to the graph of $y=x+x^{1/2}$ at point .

Tap to reveal answer

Answer

$y=x+x^{1/2}$

$y'=1+\frac{1}{2}x^{-1/2}$

$y'=1+\frac{1}{2(x^{1/2})}$

Now we plug in .

$y'=1+\frac{1}{2(4^{1/2})} = 1+\frac{1}{4} = \frac{5}{4}$ this is the slope.

Now use the point slope formula to find the tangent line.

$y-6=\frac{5}{4}(x-4)$

$y-6=\frac{5}{4}x-5$

$y=\frac{5}{4}x+1$

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Question

Practicing the chain rule level 1 A!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

can be treated as a composition of the functions

and .

in terms of and is actually , which means

since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

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Question

Practicing the chain rule level 1 B!

Find the derivative of the function

Tap to reveal answer

Answer

To understand why the answer is

,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

can be treated as a composition of the functions

and .

in terms of and is actually , which means

since in is substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the component composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

is

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

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