Concept of the Derivative - AP Calculus AB

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

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Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

← Didn't Know|Knew It →

Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

← Didn't Know|Knew It →

Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

← Didn't Know|Knew It →

Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

← Didn't Know|Knew It →

Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

← Didn't Know|Knew It →

Question

Evaluate:

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

The notation is alluding to the fact that the limit is a function of , not necessarily a "number."

Tap to reveal answer

Answer

$L(x)=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

We are not going to evaluate this directly. Looking at the function, it seems very similar to the definition of a derivative of some kind of trigonometric function. We will look at the definition of a derivative to find a function that would have a derivative defined by this limit.

In otherwords, we wish to identify a function such that its' derivative is the function .

Let's find such that:

$f'(x)=\lim_{h->0}\frac{f(x+h)-f(h)}{h}=\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h}$

Compare corresponding terms in the numerators in the above expressions.

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}$

$f(h) = \sin\left \{ h^2+h\right \}$

By inspection, these terms clearly indicate that our function must be of the form:

$f(x)=\sin(x^2 +x)+C$

Side note

If your confused by the inclusion of the arbitrary constant , note that when we differentiate this function the constant term will vanish since the derivative of a constant is zero. You can also see that in applying the definition of a derivative, if I included the constant in the two terms in the numerator,

$f(x+h)=\sin\left \{ (x+h)^2+(x+h) \right \}+C$

$f(h) = \sin\left \{ h^2+h\right \}+C$

the constant "C" would vanish when we subtract the latter from the former, .

Therefore, even if you didn't consider the constant when working out the function, it would not have changed the result

Because we know that

,

Simply differentiate $f(x)=\sin(x^2+x)$ to find ,

$f'(x)=\underset{Chain Rule}{ \underbrace{(2x+1)}}\cos(x^2+x)$

Therefore,

$L(x)=(2x+1)\cos(x^2+x)$

Or to put it another way,

$\lim_{h->0}\frac{\sin\left \{(x+h)^2 +(x+h) \right \} -\sin\left \{h^2+h \right \}}{h} =(2x+1)\cos(x^2+x)$

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Question

Find the derivative of the following function:

Tap to reveal answer

Answer

We use the power rule on each term of the function.

The first term

becomes

$2 \times 6 x^2 = 12 x$ .

The second term

becomes

.

The final term, 7, is a constant, so its derivative is simply zero.

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Question

What is the derivative of ?

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Answer

To get $\frac{\mathrm{d} }{\mathrm{d} x}(2x)$ , we can use the power rule.

Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=12x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2x^{0}$

Anything to the power is .

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=21$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2$

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Question

What is the derivative of ?

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Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2x^{2-1})+(15x^{1-0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5$

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Question

What is the derivative of ?

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Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})+(08x^{0-1})$

Notice that $(08x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=51$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=?$

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Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})+(08x^{0-1})$

Notice that $08x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

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Question

What is the derivative of ?

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Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})+(0x^{0-1})$

Notice that $(0x^{0-1})=0$ since anything times zero is zero.

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})$ .

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})$

As stated earlier, anything to the zero power is one, leaving us with:

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1$

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Question

What is the derivative of ?

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Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})+(04x^{0-1})$

Notice that $(04x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{1})+(113x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

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Question

What is the derivative of $2x^4+\frac{1}{2}x$ ?

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Answer

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{4-1})+(1\frac{1}{2}x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

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Question

What is the derivative of ?

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Answer

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3$

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