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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #31 : Derivative Review

What is the derivative of the function $f(x)=e^{\sin e^x}$ ?

Possible Answers:

$f'(x) = e^xe^{\sin e^x}$

$f'(x) = (\cos e^x) e^xe^{\sin e^x}$

$f'(x) =e^{\cos e^x}$

$f'(x) = (\sin e^x) e^{\sin e^x}$

Correct answer:

$f'(x) = (\cos e^x) e^xe^{\sin e^x}$

Explanation:

With an application of the chain rule, which is

f(x)=g(h(x))

$\Rightarrow f'(x)=g'(h(x))h'(x)$

we get that the derivative of $f(x)=e^{\sin e^x}$ is

$f'(x) = (\cos e^x) e^xe^{\sin e^x}$

since g(x)=e^x and $h(x)=\sin e^x$ which mean that

g'(x)=e^x

and

$h'(x)=e^x\cos e^x$ .

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Example Question #32 : Definition Of Derivative

Using the limit definition of a derivative, find the acceleration of the particle at x=1 if its velocity is given by the following function:

f(x)=x^2+9x

Possible Answers:

Correct answer:

Explanation:

The limit definition of a derivative is

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

where h represents a very small change in x.

The acceleration of the particle is given by the derivative of the velocity function.

So, use the given function and the above formula to write out the limit:

$\lim_{h\rightarrow 0}\frac{(x+h)^2+9(x+h)-(x^2+9x)}{h}$

which simplified becomes

$\lim_{h\rightarrow 0} \frac{h(2x+9)}{h}=2x+9$ .

Now, given the point x=1, we get an acceleration of 2(1)+9=11 .

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Example Question #1 : Derivative Rules For Sums, Products, And Quotients

Compute the derivative:

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

Possible Answers:

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

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

$f'(x)=\frac{6x^2}{1+3x^2}+4x\tan^{-1}(3x)+\frac{100}{x}$

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

$f'(x)=\frac{2x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{50}{x}$

Correct answer:

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

Explanation:

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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Example Question #32 : Derivative Review

Find dy/dx using Logarithmic Differentiation:

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

Possible Answers:

$\frac{(3x+3)^5\ln(5x)}{\tan^{-1}(2x)}[\frac{15}{3x+1}+\frac{1}{x\ln(5x)}-\frac{2}{(1+4x^2)\tan^{-1}(2x)}]$

$\frac{(3x+3)^5\ln(5x)}{\tan^{-1}(2x)}[\frac{5}{x+1}+\frac{1}{x\ln(5x)}-\frac{2}{(1+4x^2)\tan^{-1}(2x)}]$

$\frac{(3x+3)^5\ln(5x)}{\tan^{-1}(2x)}[\frac{1}{x+1}+\frac{1}{x\ln(5x)}-\frac{1}{(1+4x^2)\tan^{-1}(2x)}]$

$[\frac{1}{x+1}+\frac{1}{x\ln(5x)}-\frac{1}{(1+4x^2)\tan^{-1}(2x)}]$

$[\frac{5}{x+1}+\frac{1}{x\ln(5x)}-\frac{2}{(1+4x^2)\tan^{-1}(2x)}]$

Correct answer:

$\frac{(3x+3)^5\ln(5x)}{\tan^{-1}(2x)}[\frac{5}{x+1}+\frac{1}{x\ln(5x)}-\frac{2}{(1+4x^2)\tan^{-1}(2x)}]$

Explanation:

Evaluating the derivative using logarithmic differentiation requires us to take the natural logarithm of both sides of the equation:

$\ln (y)=\ln(\frac{(3x+3)^5\ln(5x)}{\tan^{-1}(2x)}) \rightarrow (1)$

Now, by using the rules of logarithms:

$\ln(ab)=\ln(a)+\ln(b)$

$\ln(\frac{a}{b})=\ln(a)-\ln(b)$

$\ln(a)^b=b\ln(a)$

We can rewrite the above expression (1), as:

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

This becomes:

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

If now take the derivative of each term with respect to x, we will get:

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

Simplifying further:

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

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

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

We can solve for dy/dx by multiplying both sides by y, and replacing it with the original expression.

$\frac{dy}{dx}=y[\frac{5}{x+1}+\frac{1}{x\ln(5x)}-\frac{2}{(1+4x^2)\tan^{-1}(2x)}]$

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

This is one of the answer choices.

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Example Question #31 : Derivative Review

Find dy/dx:

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

Possible Answers:

$\frac{dy}{dx}=2(x+1)^{10}[\frac{11x+1}{1+4x^2(x+1)^{12}}]$

$\frac{dy}{dx}=2(x+1)^{10}[\frac{10x+1}{1+4x^2(x+1)^{2}}]$

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

$\frac{dy}{dx}=2(x+1)^9[\frac{11x+1}{1+4x^2(x+1)^{12}}]$

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

Correct answer:

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

Explanation:

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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Example Question #1161 : Calculus Ii

Find dy/dx:

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

Possible Answers:

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

$\frac{dy}{dx}=\frac{2x}{\tan^{-1}(2x^2)}$

$\frac{dy}{dx}=\frac{2x^2}{1+4x^2}$

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

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

Correct answer:

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

Explanation:

Computing the derivative for this function requires us to know the following rules:

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

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

We can now take the derivative of the given function:

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

u_1=2x^2 , $\frac{du_1}{dx}=4x$

$u_2=\tan^{-1}(u_1)$ , $\frac{du_2}{dx}=\frac{1}{1+u_1^2}\frac{du_1}{dx}$

$\frac{d}{dx}[\ln(u_2)]=\frac{1}{u_2}\frac{du_2}{dx}$

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

This simplifies to:

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

This is one of the answer choices.

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Example Question #31 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=3x+7 .

Possible Answers:

f'(x)=3

f'(x)=7

f'(x)=-3x

f'(x)=-3

f'(x)=3x

Correct answer:

f'(x)=3

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=3x+7 , we can set a=0 , calculate f(a)=f(0)=3(0)+7=7 and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(3x+7)-7}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{3x}{x}$

$f'(x)=\lim_{x\rightarrow 0}3$

f'(x)=3

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Example Question #34 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=7x-6 .

Possible Answers:

f'(x)=7x

f'(x)=7

f'(x)=-7x

f'(x)=-7

f'(x)=-6

Correct answer:

f'(x)=7

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=7x-6 , we can set a=0 , calculate f(a)=f(0)=7(0)-6=-6 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(7x-6)-(-6)}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{7x}{x}$

$f'(x)=\lim_{x\rightarrow 0}7$

f'(x)=7

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Example Question #35 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=9-5x .

Possible Answers:

f'(x)=-5x

f'(x)=9

f'(x)=5

f'(x)=-5

f'(x)=5x

Correct answer:

f'(x)=-5

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=9-5x , we can set a=0 , calculate f(a)=f(0)=9-5(0)=9 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(9-5x)-(9)}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}-\frac{5x}{x}$

$f'(x)=\lim_{x\rightarrow 0}-5$

f'(x)=-5

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Example Question #36 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=2-11x .

Possible Answers:

Correct answer:

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=2-11x , we can set a=0 , calculate f(a)=f(0)=2-11(0)=2 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(2-11x)-2}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{-11x}{x}$

$f'(x)=\lim_{x\rightarrow 0}-11$

f'(x)=-11

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #31 : Derivative Review

Example Question #32 : Definition Of Derivative

Example Question #1 : Derivative Rules For Sums, Products, And Quotients

Compute the derivative:

Example Question #32 : Derivative Review

Find dy/dx using Logarithmic Differentiation:

Example Question #31 : Derivative Review

Find dy/dx:

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

Example Question #1161 : Calculus Ii

Find dy/dx:

Example Question #31 : Definition Of Derivative

Example Question #34 : Definition Of Derivative

Example Question #35 : Definition Of Derivative

Example Question #36 : Definition Of Derivative

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