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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #26 : Definition Of Derivative

Given f(x)=x^3+x^2+1 , what is f'(1) ?

Possible Answers:

f'(1)=1

f'(1)=6

f'(1)=5

f'(1)=0

Correct answer:

f'(1)=5

Explanation:

We can find the derivative of f(x)=x^3+x^2+1 and simply plug in x=1 to get f'(1) :

To find the derivative of this function we will need to use the power rule which states,

$x^n \ dx\rightarrow nx^{n-1}$ .

Applying the power rule we get the following.

$f'(x)=3x^{3-1}+2x^{2-1}+0$

f'(x)=3x^2+2x

so then we get

$f'(1)=3\cdot 1^2+2\cdot 1=5$ .

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Example Question #27 : Definition Of Derivative

Given $f(x)=e^{x^2+1}$ , what is f'(5) ?

Possible Answers:

$f'(5)=e^{26}$

$f'(5)=5e^{26}$

$f'(5)=10e^{26}$

$f'(5)=10e^{13}$

Correct answer:

$f'(5)=10e^{26}$

Explanation:

We can find the derivative of $f(x)=e^{x^2+1}$ and simply plug in x=5 to get f'(5) :

To find the derivative of this function we will need to use the power rule, chain rule, and rule of exponentials.

Power Rule:

$x^n \ dx=nx^{n-1}$

Chain Rule:

$f(g(x)) \ dx=f'(g(x))\cdot g'(x)$

Rule of Exponentials:

$e^u \ dx=e^u \frac{du}{dx}$

Applying these rules we get the following.

$f'(x)=e^{x^2+1}\cdot (x^2+1)dx$

$f'(x)=e^{x^2+1}\cdot (2x^{2-1}+0)$

$f'(x)=2xe^{x^2+1}$

so then we get

$f'(5)=10e^{25+1}=10e^{26}$ .

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Example Question #28 : Definition Of Derivative

What is the derivative of $f(x)=e^{x^3}$ ?

Possible Answers:

$f'(x) = 3e^{x^3}$

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

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

$f'(x) = 3xe^{x^3}$

Correct answer:

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

Explanation:

With a simple application of the chain rule, which is

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

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

with g(x)=e^x and h(x)=x^3 we get that the derivative of $f(x)=e^{x^3}$ is

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

since h'(x)=3x^2 and g'(x)=e^x .

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Example Question #29 : Definition Of Derivative

What is the derivative of the function $f(x)=e^{x^2+x}$ ?

Possible Answers:

$f'(x) = (2x+1)e^{x^2+x}$

$f'(x) = e^{x^2+x}$

$f'(x) = 2xe^{x^2+x}$

$f'(x) = (2x+1)e^{2x+1}$

$f'(x) = e^{2x+1}$

Correct answer:

$f'(x) = (2x+1)e^{x^2+x}$

Explanation:

With a simple 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 $f(x)=e^{x^2+x}$ is

$f'(x) = (2x+1)e^{x^2+x}$

since h'(x)=2x+1 .

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Example Question #30 : Definition Of Derivative

What is the derivative of the function

$f(x)=e^{\cos x^2}$ ?

Possible Answers:

$f'(x) = 2xe^{\sin x^2}$

$f'(x) = e^{-\sin 2x}$

$f'(x) = e^{\cos 2x}$

$f'(x) = -2x\sin x^2 e^{\cos x^2}$

Correct answer:

$f'(x) = -2x\sin x^2 e^{\cos x^2}$

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 $f(x)=e^{\cos x^2}$ is

$f'(x) = -2x\sin x^2 e^{\cos x^2}$

Because we have that g(x)=e^x and $h(x)=\cos x^2$ which means

g'(x)=e^x

$h'(x)=-2x\sin x^2$ .

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Example Question #31 : Definition Of Derivative

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

Possible Answers:

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

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

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

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

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:

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

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

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

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

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

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 #391 : Ap Calculus Bc

Find dy/dx:

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

Possible Answers:

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

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

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

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

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #26 : Definition Of Derivative

Example Question #27 : Definition Of Derivative

Example Question #28 : Definition Of Derivative

Example Question #29 : Definition Of Derivative

Example Question #30 : Definition Of Derivative

Example Question #31 : Definition Of Derivative

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 #391 : Ap Calculus Bc

Find dy/dx:

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

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