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Calculus 2 : Calculus II

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

Example Questions

Example Question #1 : First And Second Derivatives Of Functions

Determine the derivative of y^2=x^2+5y with the respect to .

Possible Answers:

$\frac{dy}{dx}=\frac{2x}{2y-5}$

$\frac{dy}{dx}=\frac{2x+5}{2}$

$\frac{dy}{dx}=\frac{2x+5}{2y}$

$\frac{dy}{dx}=2y=2x+5$

$\frac{dy}{dx}=\frac{2x-5}{2y}$

Correct answer:

$\frac{dy}{dx}=\frac{2x}{2y-5}$

Explanation:

In order so solve the derivative with the respect to x, implicit differentiation is required. The notation for finding the derivative of the function with the respect to x is:

$\frac{dy}{dx}$

Take the derivative and apply chain rule where necessary.

y^2=x^2+5y

$2y \cdot \frac{dy}{dx}=2x +5 \cdot \frac{dy}{dx}$

$2y \cdot \frac{dy}{dx}-5 \cdot \frac{dy}{dx}=2x$

$\frac{dy}{dx}[2y-5]=2x$

$\frac{dy}{dx}=\frac{2x}{2y-5}$

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Example Question #214 : Calculus 3

Find the derivative of $\small lnx^{lnx}$ .

Possible Answers:

$\small lnxlnx/x$

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

$\small \small lnx^{lnx}(ln(lnx)+1/x)$

$\small \small lnx^{lnx}ln(lnx)/x$

$\small (lnxlnx^{lnx-1})/x$

Correct answer:

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

Explanation:

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\small lna^{b}=blna$ to solve an easier derivative.

Let $\small y=lnx^{lnx}$ .

Now we'll take the natural log of both sides to get

$\small lny=ln(lnx^{lnx})=lnxln(lnx)$ .

Now we can use implicit differentiation to solve for $\small y'$ .

The derivative of $\small lny$ is $\small y'/y$ , and the derivative of $\small \small \small lnx(ln(lnx))$ can be found using the product rule, which states

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'$ where $\small u$ and $\small v$ are functions of $\small x$ .

Letting $\small u=lnx$ and $\small v=ln(lnx)$

(which means $\small u'=1/x$ and $\small v'=1/(xlnx)$ ) we get our derivative to be $\small ln(lnx)/x+1/x$ .

Now we have $\small y'/y=ln(lnx)/x+1/x$ , but $\small y=lnx^{lnx}$ , so subbing that in we get

$\small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x$ .

Multiplying both sides by $\small lnx^{lnx}$ , we get

$\small y'=lnx^{lnx}(ln(lnx)/x+1/x)$ .

That is our derivative.

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Example Question #1 : First And Second Derivatives Of Functions

Find the first derivative of the function:

$f(x)=\cos(x^3)+\sqrt{x^2+1}$

Possible Answers:

$-3x^2\sin(x^3)+{x}{\sqrt{x^2+1}}$

$-\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

$3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

$-3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

Correct answer:

$-3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

Explanation:

The derivative of the function is equal to

$-3x^2\sin(x^3)+\frac{x}{\sqrt{x^2+1}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

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Example Question #9 : First And Second Derivatives Of Functions

Find the second derivative of the following function:

$f(x)=2xe^{3x}+\tan(x)$

Possible Answers:

$12e^{3x}+18xe^{3x}+2\sec(x)\tan(x)$

$6e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

$18xe^{3x}+2\sec^2(x)\tan(x)$

$12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

Correct answer:

$12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

Explanation:

The first derivative of the function is equal to

$f'(x)=2e^{3x}+6xe^{3x}+\sec^2(x)$

The second derivative of the function (the derivative of the above function) is

$f''(x)=12e^{3x}+18xe^{3x}+2\sec^2(x)\tan(x)$

The following rules were used for the derivatives:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

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Example Question #3 : First And Second Derivatives Of Functions

The position of a car is given by the following function:

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Example Question #211 : Derivative Review

Find the derivative of the following function:

$f(x)=\sec(3x)+e^{2x^2+1}$

Possible Answers:

$3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

$3\sec(3x)\tan(3x)-e^{2x^2+1}$

$\sec(3x)\tan(3x)+4xe^{2x^2+1}$

$3\sec(x)\tan(x)+4xe^{2x^2+1}$

$3\sec(3x)\tan(3x)+e^{2x^2+1}$

Correct answer:

$3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

Explanation:

The derivative of the function is equal to

$f'(x)=3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

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Example Question #1 : Velocity, Speed, Acceleration

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=\pi$

$f''(x)=\0$

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

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

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

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

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

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

Correct answer:

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

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

Explanation:

In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derviative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[e^u]$

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

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

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

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

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

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Example Question #212 : Derivative Review

Find the derivative of the following function:

$f(x)=\cos(e^{3x})+\tan(2x)$

Possible Answers:

$3e^{3x}\sin(e^{3x})-\sec^2(2x)$

$-\sin(e^{3x})+\sec^2(2x)$

$e^{3x}\sin(e^{3x})+2\sec^2(2x)$

$-3e^{3x}\sin(e^{3x})+2\sec^2(2x)$

Correct answer:

$-3e^{3x}\sin(e^{3x})+2\sec^2(2x)$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

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Example Question #213 : Derivative Review

Find the derivative of the function:

$f(x)=\frac{1}{3x\sqrt{x}}+\sec(x)$

Possible Answers:

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)+\tan(x) \right |+C$

$-\frac{6}{\sqrt{x}}+\sec(x)\tan(x)+C$

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)\tan(x) \right |+C$

$\frac{-3}{2\sqrt{x}}+\sec(x)\tan(x)$

$\frac{-6}{\sqrt{x}}-\ln\left |\sec(x)+\tan(x) \right |+C$

Correct answer:

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)+\tan(x) \right |+C$

Explanation:

The integral of the function is equal to

$\frac{-6}{\sqrt{x}}+\ln\left |\sec(x)+\tan(x) \right |+C$

and was found using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int \sec(x)dx=\ln\left | \sec(x)+\tan(x)\right |+C$

Note that it is easier to integrate the first term once it is rewritten as $3x^{-\frac{3}{2}}$ .

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Example Question #211 : Derivatives

Find the velocity function of the particle if its position is given by the following function:

$s(t)=e^t\tan(t)+t^5+3$

Possible Answers:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

Correct answer:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Explanation:

The velocity function is given by the first derivative of the position function:

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : First And Second Derivatives Of Functions

Example Question #214 : Calculus 3

Example Question #1 : First And Second Derivatives Of Functions

Example Question #9 : First And Second Derivatives Of Functions

Example Question #3 : First And Second Derivatives Of Functions

Example Question #211 : Derivative Review

Example Question #1 : Velocity, Speed, Acceleration

Example Question #212 : Derivative Review

Example Question #213 : Derivative Review

Example Question #211 : Derivatives

All Calculus 2 Resources

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