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AP Calculus AB : Derivatives

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Example Questions

Example Question #416 : Ap Calculus Ab

Differentiate $f(x)=e^{3x^2-4x+3}$

Possible Answers:

$(6x-4)e^{3x^2-4x+3}$

$e^{3x^2-4x+3}$

$(6x-4)e^{6x-4}$

$(6x)e^{3x^2-4x+3}$

$e^{6x-4}$

Correct answer:

$(6x-4)e^{3x^2-4x+3}$

Explanation:

In this chain rule derivative, we need to remember what the derivative of the base function ( e^x ). The exponential function is a special function that always returns the original function. In the chain rule application, we just need to multiply that by the derivative of the exponent.

e^u'=e^u*u'

$e^{3x^2-4x+3}'=(6x-4)e^{3x^2-4x+3}$ .

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Example Question #417 : Ap Calculus Ab

Find the derivative of $f(x)=tan^{2}x$ .

Possible Answers:

$2tan^{2}xsecx$

$2sec^{2}x$

$2tanxsec^{2}x$

$sec^{2}x$

2tanx

Correct answer:

$2tanxsec^{2}x$

Explanation:

Be careful with this derivative because it is a hidden chain rule! Let's start with rewriting the problem:

$tan^{2}x=(tanx)^2.$

Now, it is easier to notice that it is a chain rule. A chain rule involves differentiating the outermost function and then, multiplying by the derivative of the inner part.

$(tanx)^2=2tanx^1*tanx'=2tanx*sec^{2}x.$

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Example Question #201 : Derivatives

Compute the derivative of f(x)=sin(cos(2x)).

Possible Answers:

-2cos(sin(2x))

2sin(2x)cos(cos(2x))

-2cos^2(2x)

-2sin(2x)cos(cos(2x))

2cos(cos(2x))

Correct answer:

-2sin(2x)cos(cos(2x))

Explanation:

This is a multiple chain derivative. With chain derivatives, we always want to start on the outermost function (keeping the rest the same), and then, multiply by the inner function derivative until we have take the derivative of every part.

sin(cos(2x))'=cos(cos(2x))*-sin(2x)*2

=-2sin(2x)cos(cos(2x)).

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Example Question #201 : Derivatives

Find the derivative: $\frac{dy}{dx}$

$x\sin(y)+y^2x=4x$

Possible Answers:

$\frac{dy}{dx}=\frac{1-xy^2+\cos(y)}{y\sin(y)+1}$

$\frac{dy}{dx}=\frac{y-\cos(y)}{x\sin(y)+y}$

$\frac{dy}{dx}=\frac{2+yx^2+\sin(y)}{y\cos(y)+1}$

$\frac{dy}{dx}=\frac{1-xy^2+x\cos(y)}{x\sin(y)+y}$

$\frac{dy}{dx}=\frac{4-y^2-\sin(y)}{x\cos(y)+2xy}$

Correct answer:

$\frac{dy}{dx}=\frac{4-y^2-\sin(y)}{x\cos(y)+2xy}$

Explanation:

$x\sin(y)+y^2x=4x$

Implicit differentiation is useful when we have an equation containing a dependent variable , and and an independent variable , but we cannot solve to write explicitly in terms of .

Differentiate both sides simultaneously:

$\underset{Product\: \: Rule \: \: 1st\: \: term}{ \underbrace{\sin(y)\left(\frac{d}{dx}x \right )+x\left(\frac{d}{dx}\sin (y) \right ) }} + \underset{Product\: \: Rule \: \: 2nd\: \: term}{ \underbrace{y^2\left(\frac{d}{dx}x\right)+x\left(\frac{d}{dx} y^2\right ) }}=\frac{d}{dx}4x$

The chain rule must be applied to the following derivatives:

$\frac{d}{dx}\sin (y)\: \; \; \; \; \; \; (1)\: \: \: \: \: \: \: \:\frac{d}{dx}y^2\: \: \: \: \: \: \:(2)$

Noting the is implied to be a function of , we have to first differentiate with respect to and then differentiate with respect to Since we do not have an explicit definition for , we simply express its' derivative in the equation as $\frac{dy}{dx}$ , treating it as an "unknown," which needs to be solved for algebraically at a later step.

For (1):

$\frac{d}{dx}\sin (y)=\left( \frac{d}{dy}\sin (y) \right )\frac{dy}{dx}=\left(\cos(y)\right )\frac{dy}{dx}$

For (2):

$\frac{d}{dx}y^2=\left(\frac{d}{dy}y^2\right)\frac{dy}{dx}=2y\frac{dy}{dx}$

Using these derivatives the main equation becomes:

$\sin (y)+x\cos (y)\frac{dy}{dx} +y^2 + 2xy\frac{dy}{dx}=4$

Now we simply solve for $\frac{dy}{dx}$ and simplify as much as possible. Isolate all terms with a derivative onto one side of the equation and factor:

$x\cos(y)\frac{dy}{dx}+2xy\frac{dy}{dx}=4-y^2-\sin(y)$

$\frac{dy}{dx}[x\cos(y)+2xy]=4-y^2-\sin(y)$

$\frac{dy}{dx}=\frac{4-y^2-\sin(y)}{x\cos(y)+2xy}$

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Example Question #21 : Chain Rule And Implicit Differentiation

Find the derivative of the function:

$h(x)=e^{2x^3}$

Possible Answers:

$h'(x)=e^{2x^3}$

$h'(x)=e^{2x^3}\cdot ln(e) \cdot 2x^3$

$h'(x)=e^{2x^3}\cdot 6x^2$

$h'(x)=e^{6x^2}\cdot 6x^2$

$h'(x)=e^{2x^3}\cdot 2x^3$

Correct answer:

$h'(x)=e^{2x^3}\cdot 6x^2$

Explanation:

On this problem we have to use chain rule, which is:

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

So in this problem we let

g(x)=e^x

and

f(x)=2x^3 .

Since

g'(x)=e^x

and

f'(x)=6x^2 ,

we can conclude that

$h'(x)=\frac{d}{dx}g(f(x))=e^{f(x)}\cdot 6x^2=e^{2x^3}\cdot 6x^2$

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Example Question #22 : Chain Rule And Implicit Differentiation

Find the derivative of:

$h(x)=(5-5x)^{3/2}$

Possible Answers:

$h'(x)=\frac{-15}{2}(5-5x)^{-1/2}$

$h'(x)=\frac{3}{2}(5-5x)^{1/2}$

$h'(x)=\frac{-15}{2}(5-5x)^{3/2}$

$h'(x)=\frac{3}{2}(5-5x)^{3/2}$

$h'(x)=\frac{-15}{2}(5-5x)^{1/2}$

Correct answer:

$h'(x)=\frac{-15}{2}(5-5x)^{1/2}$

Explanation:

On this problem we have to use chain rule, which is:

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

So in this problem we let

$g(x)=x^{3/2}$

and

f(x)=5-5x .

Since

$g'(x)=\frac{3}{2}x^{1/2}$

and

f'(x)=-5 ,

we can conclude that

$h'(x)=\frac{d}{dx}g(f(x))=\frac{3}{2}(f(x))^{1/2}\cdot -5$

$=\frac{3}{2}(5-5x)^{1/2}\cdot -5=\frac{-15}{2}(5-5x)^{1/2}$

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Example Question #121 : Computation Of The Derivative

Find the derivative of:

$k(t)=\sin(\cos t)$

Possible Answers:

$k'(t)=-\cos(\cos t)\cdot(-\sin t)$

$k'(t)=\cos(\cos t)$

$k'(t)=\sec(\cos t)\cdot(-\sin t)$

$k'(t)=\cos(\cos t)\cdot(-\sin t)$

$k'(t)=\cos(\cos t)\cdot \sin t$

Correct answer:

$k'(t)=\cos(\cos t)\cdot(-\sin t)$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

$g(t)=\sin t$

and

$f(t)=\cos t$ .

Since

$g'(t)=\cos t$

and

$f'(t)=-\sin t$ ,

we can conclude that

$k'(t)=\frac{d}{dt}g(f(t))=\cos(f(t))\cdot(-\sin t)$

and

$k'(t)=\cos(\cos t)\cdot(-\sin t)$

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Example Question #23 : Chain Rule And Implicit Differentiation

Find the derivative of:

$f(x)=\cos(1+\tan2x)$

Possible Answers:

$f'(x)=-2\sin(1+\tan 2x)(1+\sec^2(2x))$

$f'(x)=-2\sin(1+\tan 2x)\sec(2x)\tan(2x)$

$f'(x)=-\sin(1+\tan 2x)\sec^2(2x)$

$f'(x)=-2\sin(1+\tan 2x)\sec^2(2x)$

$f'(x)=-\sin(1+\tan 2x)$

Correct answer:

$f'(x)=-2\sin(1+\tan 2x)\sec^2(2x)$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(h(j(x)))=g'(h(j(x)))h'(j(x))j'(x)$

So in this problem we let

$g(x)=\cos x$

and

$h(x)=1+\tan x$

and

j(x)=2x .

Since

$g'(x)=-\sin x$

and

$h'(x)=\sec^2(x)$

and

j'(x)=2 ,

we can conclude that

$f'(x)=\frac{d}{dx}g(h(j(x)))=-sin(h(j(x)))\cdot (sec^2(j(x)))\cdot (-2)$

$f'(x)=-2\sin(1+\tan2x)(\sec^2(2x))$

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Example Question #24 : Chain Rule And Implicit Differentiation

Find the derivative of the function:

$h(t)=(\ln t)^3$

Possible Answers:

$h'(t)=\frac{3(\ln t)^2}{t^2}$

$h'(t)=\frac{3(\ln t)^2}{t}$

$h'(t)=3(\ln t)^2\cdot t$

$h'(t)=\frac{-3(\ln t)^2}{t}$

$h'(t)=\3(\ln t)^2$

Correct answer:

$h'(t)=\frac{3(\ln t)^2}{t}$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

g(t)=t^3

and

$f(t)=\ln t$ .

Since

g'(t)=3t^2

and

$f'(t)=\frac{1}{t}$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=3(f(t))^2\cdot\frac{1}{t}$

and

$h'(t)= \frac{3(\ln t)^2}{t}$

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Example Question #25 : Chain Rule And Implicit Differentiation

Find the derivative of:

$n(x)=5^{7x-3}$

Possible Answers:

$n'(x)=5^{7x-3}\ln(7x-3)$

$n'(x)=5^{7x-3}\ln(5)$

$n'(x)=5^{7x-3}\ln(7x-3)\cdot 7$

$n'(x)=(7x-3)\ln(5)\cdot 7$

$n'(x)=5^{7x-3}\ln(5)\cdot 7$

Correct answer:

$n'(x)=5^{7x-3}\ln(5)\cdot 7$

Explanation:

On this problem we have to use chain rule, which is:

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

So in this problem we let

g(x)=5^x

and

f(x)=7x-3 .

Since

$g'(x)=5^x\ln(5)$

and

f'(x)=7 ,

we can conclude that

$n'(x)=\frac{d}{dx}g(f(x))=5^{f(x)}\ln(5)\cdot 7=5^{7x-3}\ln(5)\cdot 7$

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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 #416 : Ap Calculus Ab

Example Question #417 : Ap Calculus Ab

Example Question #201 : Derivatives

Example Question #201 : Derivatives

Example Question #21 : Chain Rule And Implicit Differentiation

Example Question #22 : Chain Rule And Implicit Differentiation

Example Question #121 : Computation Of The Derivative

Example Question #23 : Chain Rule And Implicit Differentiation

Example Question #24 : Chain Rule And Implicit Differentiation

Example Question #25 : Chain Rule And Implicit Differentiation

All AP Calculus AB Resources

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