Derivative rules for sums, products, and quotients of functions - AP Calculus AB

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Question

$\begin{align}&\text{Take the derivative with respect to }x\text{ of the function}\&y=-\frac{(380tan(17x))}{2^{(20x)}}\end{align}$

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Answer

$\begin{align}&\text{The function that we have been given can be seen as two smaller functions}\&\text{multiplied together, the first being: }tan(17x)\&\text{and the second being: }\frac{1}{2^{(20x)}}\&\text{With the product scaled by }-380\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\&\text{along with the following:}\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\&d[b^{ax}]=ab^{ax}ln(b)sx\&\text{From the above we can continue:}\&u=tan(17x);du=17tan(17x)^{2} + 17\&v=\frac{1}{2^{(20x)}};dv=-\frac{(20ln(2))}{2^{(20x)}}\&\text{We can then combine these terms (Don't forget the scalar!)}\&\text{So the derivative is:}\&\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}\end{align}$

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Question

Find the derivative of the following function:

$f=\frac{4x^3+\sin(x)}{\tan(x)}$

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Answer

To find the derivative of the function, we use the quotient rule, which is

$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{f}{g})=\frac{gf'-fg'}{g^2}$ , where and are any expression

Using the function from the problem statement, we get

$f'=\frac{\tan(x)\frac{\mathrm{d} }{\mathrm{d} x}(4x^3+\sin(x))-(4x^3+\sin(x))\frac{\mathrm{d} }{\mathrm{d} x}(\tan(x))}{\tan^2(x)}$

Taking the derivatives, we get

$f'=\frac{\tan(x)(12x^2+\cos(x))-(4x^3+\sin(x))(\sec^2(x))}{\tan^2(x)}$

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Question

Find the slope of the $f(x)=\sqrt{x^2 +2x-3}$ at .

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Answer

First we need to find the derivative of the function. $f'(x)=\frac{1+x}{\sqrt{-3+2x+x^2}}$

Now, we can plug in to the derivative function.

$f'(2)=\frac{3\sqrt{5}}{5}$

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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.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2$

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Question

Find the derivative of the function,

$y =x^2\sin(3x)$

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Answer

$y =x^2\sin(3x)$

Differentiate both sides and proceed with the product rule:

$\frac{dy}{dx}=\frac{d}{dx}[x^2\sin(3x)]$ (1)

$\frac{dy}{dx}= x^2\frac{d}{dx}\sin(3x)+ \sin(3x)\frac{d}{dx}x^2$

Evaluate the derivatives in each term. For the first term,

$=x^2\underbrace{\frac{d}{dx}\sin(3x)} + \sin(3x)\frac{d}{dx}x^2$ (2)

apply the chain rule,

$\frac{d}{dx}sin(3x)=3\cos(3x)$

So now the first term in equation (2) can be written,

$x^2\frac{d}{dx}\sin(3x)=x^2[3\cos(3x)]=3x^2\cos(3x)$ (3)

The second term in equation (2) is easy, this is just the product of $\sin(3x)$ multiplied by the derivative of ,

$\sin(3x)\frac{d}{dx}x^2 = 2x\sin(3x)$ (4)

Combine equations (3) and (4) to write the derivative,

$\frac{dy}{dx} =3x^2\cos(3x)+2x\sin(3x)$

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Question

Find the derivative.

$\sin (x)x^2$

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Answer

Use the product rule to find the derivative.

$\cos (x)x^2+2x\sin (x)$

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Question

Find the derivative.

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Answer

Use the power rule to find the derivative.

$\frac{d}{dx}x^3=3x^2$

$\frac{d}{dx}4x^2=8x$

$\frac{d}{dx}-x=-1$

Thus, the derivative is

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Question

Find given $f(x)=3sin(x^2)*e^{2x-1}$

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Answer

Here we use the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}(F(x)\cdot G(x))=F'(x)\cdot G(x)+F(x)\cdot G'(x)$

Let and $G(x)=e^{2x-1}$

Then $F'(x)=3cos(x^2)\cdot2x$ (using the chain rule)

and $G'(x)=e^{2x-1}\cdot2$ (using the chain rule)

Subbing these values back into our equation gives us

$f'(x)=3cos(x^2)\cdot2x\cdot e^{2x-1}+3sin(x^2)\cdot e^{2x-1}\cdot2$

Simplify by combining like-terms

$f'(x)=6xcos(x^2)\cdot e^{2x-1}+6sin(x^2)\cdot e^{2x-1}$

and pulling out a $6e^{2x-1}$ from each term gives our final answer

$f'(x)=6e^{2x-1}(xcos(x^2)+sin(x^2))$

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Question

If $f(x)=\pi\sin(ex)$ , evaluate .

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Answer

When evaluating the derivative, pay attention to the fact that $e,\pi$ are constants, (not variables) and are treated as such.

$f'(x)=\pi\cos(ex)\times (ex)'=e\pi\cos(ex)$ .

and hence

$f'(0)=e\pi\cos(e(0))=e\pi$ .

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Question

If , evaluate

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Answer

To obtain an expression for , we can take the derivative of using the sum rule.

.

Substituting into this equation gives us

.

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Question

If $f(x)=\frac{x\tan^{-1}(x)}{\ln(x)}$ , find .

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Answer

To find , we will need to use the quotient rule; $\frac{gf'-fg'}{g^2}$ .

$f(x)= \frac{x\tan^{-1}x}{\ln(x)}$ . Start

$f'(x)= \frac{(\ln(x))(x\tan^{-1}x)'-(x\tan^{-1}x)(\ln(x))'}{(\ln(x))^2}$ . Use the quotient rule.

$= \frac{(\ln(x))(x(\frac{1}{1+x^2})+1(\tan^{-1}x))-(x\tan^{-1}x)(\frac{1}{x})}{(\ln(x))^2}$ . Take the derivatives inside of the quotient rule. The derivative of $x\tan^{-1}x$ uses the product rule.

$=\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$ . Simplify to match the correct answer.

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Question

Find the derivative of the following equation:

$h(x)=\frac{6+x^{3}}{x}$

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Answer

Given that there are 2 terms in the numerator and only one in the denominator, one can split up the equation into 2 separate derivatives:

$\frac{d}{dx}(\frac{6}{x})+\frac{d}{dx}(\frac{x^{3}}{x})$ .

Now we simplify these, and proceed to solve:

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

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Question

Find the derivative of the following equation:

$h(x) = \sin(x)\cos(x)$

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Answer

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that $\frac{d}{dx}(f(x)g(x)) = f{}'(x)g(x) + f(x)g{}'(x)$ .

By using this rule, we get the answer:

$\cos(x)\cos(x) + \sin(x)(-\sin (x))$

By simplifying, we conclude that the derivative is equal to

$\cos ^{2}(x)-\sin ^{2}(x)$ .

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Question

Find the derivative of the following equation:

$h(x)=x\ln(x)$

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Answer

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that

By applying this rule to the equation

$x\ln(x)$

we get

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

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Question

Find the derivative of the following function:

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

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Answer

Because we are dealing with a quotient that cannot be simplified, we use the quotient rule, which states that if

$h(x)=\frac{f(x)}{g(x)}$ ,

$h{}'(x)=\frac{(g(x)f{}'(x)-f(x)g{}'(x))}{(g(x))^{2}}$ .

By observing the given equation

$h(x)=\frac{x}{x^{2}+1}$ ,

we can see that

and

$g(x)=x^{2}+1$ .

Therefore, the derivative is

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

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Question

Find the derivative of the following equation:

$h(x)=\frac{x}{\ln (x)}$

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Answer

Because we are differentiating a quotient that cannot be simplified, we must use the quotient rule, which states that if

$h(x)=\frac{f(x)}{g(x)}$ ,

then

$h{}'(x)=\frac{g(x)f{}'(x)-f(x)g{}'(x)}{(g(x))^{2}}$ .

By observing the given equation,

$h(x)=\frac{x}{\ln (x)}$ ,

we see that in this case,

and

$g(x)=\ln(x)$ .

Given this information, the quotient rule tells us that

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

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Question

Find the derivative of the following equation:

$h(x)=\sin( \frac{x^{2}}{x+1})$

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Answer

This problem is a quotient rule inside of a chain rule. First, let's look at the chain rule:

.

Given this, we can deduce that since

$h(x)=\sin(\frac{x^{2}}{x+1})$ ,

$f(x)=\sin(x)$

and

$g(x)=\frac{x^{2}}{x+1}$ .

By plugging these into the chain rule formula, we get

$(f o g(x)){}'=\cos(\frac{x^{2}}{x+1})((\frac{d}{dx}(\frac{x^{2}}{x+1})))$

To find the derivative of the

second term, we must use the quotient rule, which states that the the derivative

of a quotient is ((denominator)(derivative of numerator)-(numerator)(derivative

of denominator))/(denominator squared). Using this rules we find that

$\frac{d}{dx}(\frac{x^{2}}{x+1})=\frac{((x+1)(2x)-(x^{2})(1))}{(x+1)^{2}}=\frac{x(x+2)}{(x+1)^{2}}$ .

By plugging this back in, we find the final derivative to be

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

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Question

Differentiate the function:

$y=x^2\sqrt{3x+1}$

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Answer

$y = x^2 \sqrt{3x+1}$

First notice that the function is a product of two functions and $\sqrt{3x+1}$ . Apply the product rule:

$\frac{dy}{dx}=\sqrt{3x+1}\left(\frac{d}{dx}x^2 \right )+x^2\left(\frac{d}{dx}\sqrt{3x+1} \right )$

The second term will require the chain rule. Recall that the derivative for a radical function-of-a-function $y=\sqrt{u(x)}$ is given by:

$\underset{Chain-Rule}{ \underbrace{\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}}} =\frac{1}{2\sqrt{u(x)}}u'(x)$ $\frac{dy}{dx}=\sqrt{3x+1}(2x )+x^2\left(\frac{3}{2\sqrt{3x+1}}\right)$

$=2x\sqrt{3x+1}+\frac{3x^2}{2\sqrt{3x+1}}$

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

$= \frac{4x(3x+1)+3x^2}{2\sqrt{3x+1}}$

$= \frac{12x^2+4x+3x^2}{2\sqrt{3x+1}}$

$= \frac{15x^2+4x}{2\sqrt{3x+1}}$

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Question

$\begin{align}&\text{Calculate the derivative of the function}\&-\frac{(114cos(19x))}{x^{20}}\end{align}$

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Answer

$\begin{align}&\text{The function that we have been given can be seen as two smaller functions}\&\text{multiplied together, the first being: }cos(19x)\&\text{and the second being: }\frac{1}{x^{20}}\&\text{With the product scaled by }-114\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\&\text{along with the following:}\&d[cos(ax)]=-asin(ax)dx\&d[x^a]=ax^{a-1}dx\&\text{From the above we can continue:}\&u=cos(19x);du=-19sin(19x)\&v=\frac{1}{x^{20}};dv=-\frac{20}{x^{21}}\&\text{We can then combine these terms (Don't forget the scalar!)}\&\text{So the derivative is:}\&\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}\end{align}$

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Question

$\begin{align}&\text{Calculate the derivative of the function}\&56sin(2x)tan(16x)\end{align}$

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Answer

$\begin{align}&\text{The function that we have been given can be seen as two smaller functions}\&\text{multiplied together, the first being: }tan(16x)\&\text{and the second being: }sin(2x)\&\text{With the product scaled by }56\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\&\text{along with the following:}\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\&d[sin(ax)]=acos(ax)dx\&\text{From the above we can continue:}\&u=tan(16x);du=16tan(16x)^{2} + 16\&v=sin(2x);dv=2cos(2x)\&\text{We can then combine these terms (Don't forget the scalar!)}\&\text{So the derivative is:}\&56sin(2x)\cdot (16tan(16x)^{2} + 16) + 112cos(2x)tan(16x)\end{align}$

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