Derivative Rules for Sums, Products, and Quotients - AP Calculus BC

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Question

Suppose and are differentiable functions, and . Calculate the derivative of , at

Answer

The correct answer is 11.

Taking the derivative of involves the product rule, and the chain rule.

Substituting into both sides of the derivative we get

$h'(1)= -g'(-1)f(3)+g(-1)f'(3) = -1\cdot1 + 3 \cdot 4 = 11$ .

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Question

What is the slope of at ?

Answer

In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.

Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:

Following that procedure, we set equal to and equal to .

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

which can be simplified to

.

Now plug in 1 to find the slope at x=1.

Remember that .

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Question

Find the value of the derivate of the given function at the point :

$f(x)=\frac{x^2+2}{x+4}$

Answer

To solve this problem, first we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

$\f'(x)=\frac{(x+4)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2)-(x^2+2)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+4)}{(x+4)^2}\ f'(x)=\frac{(x+4)\cdot 2x-(x^2+2)\cdot 1}{(x+4)^2}\f'(x)=\frac{2x^2+8x-x^2-2}{(x+4)^2}\f'(x)=\frac{x^2+8x-2}{(x+4)^2}$

From here we need to evaluate at the given point . In this case, only the x value is important, so we evaluate our derivative at x=2 to get

$\f'(2)=\frac{(2)^2+8(2)-2}{(2+4)^2}\ f'(2)=\frac{4+16-2}{36}\f'(2)=\frac{18}{36}=\frac{1}{2}$ .

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Question

Find the second derivative of the given function:

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

Answer

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be
$\f'(x)=\frac{x\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2x+1)-(x^2+2x+1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x)}{x^2}\f'(x)=\frac{x\cdot (2x+2)-(x^2+2x+1)\cdot 1}{x^2}\f'(x)=\frac{x^2-1}{x^2}$ .

Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.

Thus, we get

$\f''(x)=\frac{x^2\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2-1)-(x^2-1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2)}{x^4}\ f''(x)=\frac{2x^3-2x^3+2x}{x^4}\f''(x)=\frac{2x}{x^4}=\frac{2}{x^3}$

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Question

Given:

Find f'(x):

Answer

Computation of the derivative requires the use of the Product Rule and Chain Rule.

The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:

$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f(x)$

This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."

In the problem statement, we are given:

is the "First" function, and is the "Second" function.

The "Second" function requires use of the Chain Rule.

When:

$y=(f(x))^a \rightarrow \frac{dy}{dx}=a(f(x))^{a-1}*f'(x)$

Applying these formulas results in:

Simplifying the terms inside the brackets results in:

We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".

Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:

Simplifying this results in one of the answer choices:

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Question

Find the slope of the tangent line to the function $h(x)=\frac{2ln(x)}{x^2}$ at .

Answer

The slope of the tangent line to a function at a point is the value of the derivative of the function at that point. In this problem, is a quotient of two functions, $2ln(x) \ and\ x^2$ , so the quotient rule is needed.

In general, the quotient rule is

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$ .

To apply the quotient rule in this example, you must also know that $\frac{d}{dx}ln(x)=\frac{1}{x}$ and that $\frac{d}{dx}x^n=nx^{n-1}$ .

Therefore, the derivative is

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

The last step is to substitute for in the derivative, which will tell us the slope of the tangent line to at .

$h'(1)=\frac{2(1)-4(1)ln(1)}{1^4}=\frac{2-0}{1}=2$

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Question

$f(x)=x^2\tan^{-1}(3x^5)$

$\frac{dy}{dx}=?$

Answer

Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of $\tan^{-1}(x)$ .

Product Rule:

$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

Applying these two rules results in:

$\frac{dy}{dx}=\frac{d}{dx}[x^2\tan^{-1}(3x^5)]$

$\frac{dy}{dx}=(x^2)(\frac{1}{1+(3x^5)^2})(15x^4)+\tan^{-1}(3x^5)(2x)$

$\frac{dy}{dx}=\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

This matches one of the answer choices.

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Question

Compute the derivative:

$f(x)=2x^2\tan^{-1}(3x)+5\ln(10x)$

Answer

Computation of this derivative will require the use of the Product Rule, and knowledge of the derivative of the inverse tangent function, and natural logarithmic function:

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

$\frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}$

We can now easily compute the derivative.

$f'(x)=(2x^2)(\frac{1}{1+(3x)^2})(3)+(\tan^{-1}(3x))(4x)+5(\frac{1}{10x})(10)$

This simplifies to:

$f'(x)=\frac{6x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

This is one of the answer choices.

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Question

Find dy/dx:

$y=\tan^{-1}(2x(x+1)^{10})$

Answer

Solving for the derivative requires knowledge of the rule for the inverse tangent function:

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

In our case:

$u=2x(x+1)^{10}$

We can take the derivative of this using the product rule:

$\frac{du}{dx}=(2x)(10(x+1)^9(1))+(2)(x+1)^{10}=2(x+1)^9[10x+(x+1)]$

$\frac{du}{dx}=2(x+1)^9(11x+1)$

Now we can simply plug all of this into the above formula and we arrive at:

$\frac{dy}{dx}=[\frac{1}{1+(2x(x+1)^{10})^2}][2(x+1)^9(11x+1)]$

Simplifying this further gives:

$\frac{dy}{dx}=2(x+1)^{9}[\frac{11x+1}{1+4x^2(x+1)^{20}}]$

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Question

Find the derivative of the function

$f(x)= \frac{x^2}{x\ln(x)}$

Answer

We find the answer using the quotient rule

$(\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

and then simplifying.

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

$= \frac{2x^2\ln(x)-x^2-x^2\ln(x)}{x^2\ln(x)^2}$

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

$=\frac{\ln(x)-1}{\ln(x)^2}$ or $\frac{\ln(x)-1}{[\ln(x)]^2}$ . The extra brackets in the denominator are optional.

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Question

If , find in terms of and .

Answer

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

$\ln(x^y)=\ln(xy)$

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

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

$y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}$ . Quotient Rule + implicit differentiation.

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

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

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

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

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

$y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

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Question

Find the derivative of the function $y= \frac{\tan^{-1}(x)}{x}$

Answer

The correct answer is $\frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}$ .

Using the Quotient Rule and the fact $\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$ , we have

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

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

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

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

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

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

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Question

$\frac{\mathrm{d} }{\mathrm{d} x} \enspace 2 e^x x^2 -xe^x$

Answer

First, factor out : . Now we can differentiate using the product rule, .

Here, so . so .

The answer is

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Question

$\frac{\mathrm{d} }{\mathrm{d} x} 2 \sin x \cos x$

Answer

According to the product rule, . Here $f(x) = 2 \sin x$ so $f'(x) = 2 \cos x$ . $g(x) = \cos x$ so $g'(x) = -\sin x$ .

The derivative is $2 \cos x \cdot \cos x + 2 \sin x \cdot - \sin x = 2 \cos ^2 x - 2 \sin ^2 x$

Factoring out the 2 gives $2(\cos ^2 x - \sin ^2 x )$ . Remembering the double angle trigonometric identity finally gives $2 (\cos (2x ))$

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Question

If $f(t) = \frac{t}{\ln(t)}$ , find

Answer

First, we need to find . We can do that by using the quotient rule.

$f'(t) = \frac{(\ln(t))(1)-(t)(\frac{1}{t})}{(\ln(t))^2}$ .

Plugging in for and simplifying, we get

$f'(e^2) = \frac{(\ln(e^2))(1)-(e^2)(\frac{1}{e^2})}{(\ln(e^2))^2} = \frac{2-1}{2^2} = \frac{1}{4}$ .

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Question

Find the derivative of f:

$f=\frac{x^3}{x\cos(x)}$

Answer

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Question

Find the derivative of the function:

$f=x^3+x^2+\alpha^2+2\alpha^2x$

where $\alpha$ is a constant

Answer

When taking the derivative of the sum, we simply take the derivative of each component.

The derivative of the function is

$f'=3x^2+2x+2\alpha^2$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Question

Compute the first derivative of the following function.

Answer

Compute the first derivative of the following function.

To solve this problem, we need to apply the product rule:

So, we need to apply this rule to each of the terms in our function. Let's start with the first term

Next, let's tackle the second part

Now, combine the two to get:

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Question

Find the second derivative of g(x)

Answer

Find the second derivative of g(x)

To find this derivative, we need to use the product rule:

So, let's begin:

$g'(x)=12x^3ln(x)+3x^4(\frac{1}{x})=12x^3ln(x)+3x^3$

So, we are closer, but we need to derive again to get the 2nd derivative

$g''(x)=36x^2ln(x)+12x^3(\frac{1}{x})+9x^2=36x^2ln(x)+21x^2$

So, our answer is:

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Question

Evaluate the derivative of the function $y = x^2\cos(3x)$ .

Answer

Use the product rule:

where and $g(x) = \cos(3x)$ .

By the power rule, .

By the chain rule, $g'(x) = -\sin(3x)\cdot3$ .

Therefore, the derivative of the entire function is:

$y'=(2x)(\cos(3x))+(x^2)(-\sin(3x)\cdot 3)$

$y'=2x\cos(3x)-3x^2\sin(3x)$ .

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