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AP Calculus AB : Derivatives

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Example Questions

Example Question #61 : Derivative Rules For Sums, Products, And Quotients Of Functions

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

Possible Answers:

$6$

$\frac{6\sqrt{5}}{5}$

$\frac{\sqrt{5}}{5}$

$5$

$\frac{3\sqrt{5}}{5}$

Correct answer:

$\frac{3\sqrt{5}}{5}$

Explanation:

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 x=2 to the derivative function.

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

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Example Question #62 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following function:

$f=\frac{x^4\sin{x}}{x^2+3}$

Possible Answers:

$f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2-3)^2}$

$f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}$

$f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\cos(x))}{(x^2+3)^2}$

$f'=\frac{(x^2-3)(4x^3\sin(x)+x^4\cos(x)+2x^5\sin(x))}{(x^2+3)^2}$

Correct answer:

$f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}$

Explanation:

To find the derivative of the function, we use a combination of the quotient rule and product rule

$f=\frac{x^4\sin{x}}{x^2+3}$

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

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

$f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}$

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Example Question #63 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function

f=x^3+5x^2+8x+20

Possible Answers:

f'=3x^2-10x+8

f'=3x^3+10x^2+8x

$f'=\frac{3x^2+10x^2+8}{x}$

f'=3x^2+10x+8

Correct answer:

f'=3x^2+10x+8

Explanation:

To find the derivative of the function, we use the rule

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ and apply it to each term in the function

$f'=\frac{\mathrm{d} }{\mathrm{d} x}(x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(8x)+\frac{\mathrm{d} }{\mathrm{d} x}(20)$

f'=3x^2+10x+8

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Example Question #64 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following function

$f=\frac{x^3}{\tan{x}}$

Possible Answers:

$f'=\frac{3x^2\tan{x}-x^3\sec^2{x}}{\tan{x}}$

$f'=\frac{x^3\tan{x}-x^3\sec^2{x}}{\tan^2{x}}$

$f'=\frac{3x^2\tan{x}-x^3\sec^2{x}}{\tan^2{x}}$

$f'=\frac{3x^2\tan{x}+x^3\sec^2{x}}{\tan^2{x}}$

Correct answer:

$f'=\frac{3x^2\tan{x}-x^3\sec^2{x}}{\tan^2{x}}$

Explanation:

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

$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{f}{g})=\frac{gf'-fg'}{g^2}$

Applying this to the function from the problem statement, we get

$f'=\frac{(x^3)'\tan{x}-x^3(\tan{x})'}{\tan^2{x}}$

$f'=\frac{3x^2\tan{x}-x^3\sec^2{x}}{\tan^2{x}}$

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Example Question #65 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function

$f=x\sin{x}\cos{x}$

Hint: $\cos^2{x}-\sin^2{x}=\cos{2x}$

Possible Answers:

$f'=\sin{x}\cos{x}+x\cos{2x}$

$f'=\sin{x}\cos{x}+\cos{2x}$

$f'=\sin{x}\cos^2{x}+x\cos{2x}$

$f'=\sin{x}\cos{x}+x\cos{x}$

Correct answer:

$f'=\sin{x}\cos{x}+x\cos{2x}$

Explanation:

To find the derivative of the function, we use the product rule, which by definition is

$\frac{\mathrm{d} }{\mathrm{d} x}(f*g)=f'g+g'f$

In this case, we can split up the product such that

$f=(x\sin{x})*(\cos{x})$

$f'=(x\sin{x})'(\cos{x})+(x\sin{x})(\cos{x})'$

$y'=(\sin{x+x\cos{x}})(\cos{x})+\sin{x}(x\sin{x})$

$y'=\sin{x\cos{x}}+x\cos^2{x}+x\sin^2{x}$

$f'=\sin{x}\cos{x}+x\cos{2x}$

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Example Question #66 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function:

y=x^3e^x

Possible Answers:

y'=3x^2e^x

y'=-e^x(x^3+3x^2)

y'=e^x(x^3+3x^2)

$y'=\frac{e^x}{(x^3+3x^2)}$

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

Correct answer:

y'=e^x(x^3+3x^2)

Explanation:

This question requires us to understand two things.

First, the derivative of e^x is always e^x , such that the exponent contains a single variable (any other operation or numerical factors could cause the chain rule to come into play; a later topic)

Second, we must understand the product rule for derivatives. The product rule works as follows:

$\frac{dy}{dx}f(x)*g(x)=f(x)g'(x)+g(x)f'(x)$

Understanding these two concepts allows us to tackle the derivative of the given function.

y=x^3e^x

y'=(x^3)(e^x) +(e^x)(3x^2)

y'=e^x(x^3) +e^x(3x^2)

This simplifies to:

y'=e^x(x^3+3x^2)

We are finished taking the derivative of the product!

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Example Question #67 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the first and second derivatives of the function:

$r=\frac{s^3+7}{s}$

Possible Answers:

$r'=\frac{7s^3+2}{s^3}$

$r''=\frac{2s^4+14s}{s^4}$

$r'=\frac{2s^3-7}{s^2}$

$r''=\frac{2s^4+14s}{s^4}$

$r'=\frac{2s^3-7}{s^2}$

$r''=\frac{3s^4-14s}{s^4}$

$r'=-\frac{2s^3-7}{s^2}$

$r''=-\frac{2s^4+14s}{s^4}$

None of the other answers are correct

Correct answer:

$r'=\frac{2s^3-7}{s^2}$

$r''=\frac{2s^4+14s}{s^4}$

Explanation:

There are two steps to this problem:

1) Take the first derivative of the function:

$r=\frac{s^3+7}{s}$

$r'=\frac{[s(3s^2)]-[(s^3+7)(1)]}{s^2}$ (applying the quotient rule)

$r'=\frac{(3s^3)-(s^3+7)}{s^2}$

$r'=\frac{3s^3-s^3-7}{s^2}$ (simplifying and combining terms)

Thus, the first derivative is as follows:

$r'=\frac{2s^3-7}{s^2}$

2) Take the second derivative of the original function (or the first derivative of the first derivative we just found)

$r'=\frac{2s^3-7}{s^2}$

$r''=\frac{[(s^2)(6s^2)]-[(2s^3-7)(2s)]}{s^4}$ (applying the quotient rule)

$r''=\frac{(6s^4)-(4s^4-14s)}{s^4}$

$r''=\frac{6s^4-4s^4+14s}{s^4}$ (simplifying and combining terms)

Thus, our second derivative is as follows:

$r''=\frac{2s^4+14s}{s^4}$

So the derivatives of the original function "r" should be:

$r'=\frac{2s^3-7}{s^2}$

$r''=\frac{2s^4+14s}{s^4}$

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Example Question #68 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function:

$t=2s^{3/2}+3e^2$

Possible Answers:

$t'=-\sqrt{3s}$

t'=3s

$t'=-3\sqrt{s}$

$t'=\sqrt{3s}$

t'=-3s

Correct answer:

$t'=\sqrt{3s}$

Explanation:

We are given the function:

$t=2s^{3/2}+3e^2$

Applying the general power rule, we can take the derivative

$t'=\frac{6}{2}s^{1/2}+0$ (the 3e term is removed because "e" is a constant, and deriving a constant gives us zero!)

$t'=3s^{1/2}$ (further simplifying)

Thus, the derivative of our original function, and the correct answer, is:

$t'=\sqrt{3s}$

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Example Question #69 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the first derivative of the function:

$t=(1-v)(1+v^2)^{-1}$

Possible Answers:

$t'=\frac{(1+v^2)^{2}}{v^2-2v-1}$

$t'=\frac{v^2-2v-1}{(1+v^2)^{2}}$

$t'=-\frac{(1+v^2)^{2}}{v^2-2v-1}$

$t'=-\frac{v^2-2v-1}{(1+v^2)^{2}}$

$t'=-\frac{(1+v^2)^{2}}{2v^2-4v-2}$

Correct answer:

$t'=\frac{v^2-2v-1}{(1+v^2)^{2}}$

Explanation:

We are given the function:

$t=(1-v)(1+v^2)^{-1}$

The trick to making this function more approachable is to rewrite the function. The negative 1 on the right-most term allows us to rewrite the function as:

$t=\frac{1-v}{1+v^2}$

We can now apply the quotient rule for derivatives, and derive the function:

$t'=\frac{[(1+v^2)(-1)]-[(1-v)(2v)]}{(1+v^2)^{2}}$

This derivative is quite messy, but we can simplify the numerator further!

$t'=\frac{(-v^2-1)-(2v-2v^2)}{(1+v^2)^{2}}$

$t'=\frac{-v^2-1-2v+2v^2}{(1+v^2)^{2}}$

Simplifying once more, we get our answer:

$t'=\frac{v^2-2v-1}{(1+v^2)^{2}}$

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Example Question #71 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the first derivative of the function:

$t=\frac{s^2+5s-1}{s^2}$

Possible Answers:

$t'=-\frac{s^4}{15s^2-2s}$

$t'=-\frac{5s^2-8s}{s^4}$

$t'=\frac{-5s+2}{s^3}$

$2(\frac{s^4}{15s^2-2s})$

$t'=\frac{s^4}{15s^2-2s}$

Correct answer:

$t'=\frac{-5s+2}{s^3}$

Explanation:

We are given the function:

$t=\frac{s^2+5s-1}{s^2}$

As the function given to us is a quotient, we must use the quotient rule for derivatives to get the derivative of the function.

$t'=\frac{[(s^2)(2s+5)]-[(s^2+5s-1)(2s)]}{s^4}$

Technically, this derivative is correct, but it is not at all optimal. Let's clean things up and simplify our answer further:

$t'=\frac{(2s^3+5s^2)-(2s^3+10s-2s)}{s^4}$

$t'=\frac{2s^3+5s^2-2s^3-10s+2s}{s^4}$

Thus, we combine our terms, do our last simplification, and reach our answer:

$t'=\frac{-5s^2+2s}{s^4}$

$t'=\frac{-5s+2}{s^3}$

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #61 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #62 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #63 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #64 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #65 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #66 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #67 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #68 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #69 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #71 : Derivative Rules For Sums, Products, And Quotients Of Functions

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