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

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Example Question #6 : Finding Derivative Of A Function

What is the derivative of x^3+2x+5 ?

Possible Answers:

3x^2+2x

$\frac{1}{3}x^2+\frac{1}{2}x+\frac{1}{5}$

3x^2+2

Correct answer:

3x^2+2

Explanation:

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 5x^0 , as anything to the zero power is one.

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

Notice that $(0*5x^{0-1})=0$ , as anything times zero is zero.

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

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

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

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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:

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

$5$

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

$6$

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

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 #741 : Ap Calculus Ab

Find the derivative of the function

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

Possible Answers:

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

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

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

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 #751 : Ap Calculus Ab

Find the derivative of the following function

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

Possible Answers:

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

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

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

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

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 #752 : Ap Calculus Ab

Find the derivative of the function

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

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

Possible Answers:

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

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

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

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

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

Find the derivative of the function:

y=x^3e^x

Possible Answers:

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

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

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

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

y'=3x^2e^x

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 #754 : Ap Calculus Ab

Find the first and second derivatives of the function:

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

Possible Answers:

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

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

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

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

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

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

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

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

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 #755 : Ap Calculus Ab

Find the derivative of the function:

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

Possible Answers:

t'=3s

t'=-3s

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

$t'=\sqrt{3s}$

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

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 #61 : 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:

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

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

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

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

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

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

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #6 : Finding Derivative Of A Function

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 #741 : Ap Calculus Ab

Example Question #751 : Ap Calculus Ab

Example Question #752 : Ap Calculus Ab

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

Example Question #754 : Ap Calculus Ab

Example Question #755 : Ap Calculus Ab

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

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