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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 #2 : Velocity, Speed, Acceleration

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=cot(x)

f''(x)=csc^2(x)

f'(x)=csc(x)

f''(x)=-sec(x)

f'(x)=sec(x)tan(x)

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

f'(x)=sec^2(x)

f''(x)=sec^3(x)

Correct answer:

f'(x)=sec(x)tan(x)

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

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

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

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Example Question #21 : First And Second Derivatives Of Functions

Find the derivative of the following function:

$f(x)=\frac{x^3}{\sin(x)}+\ln(x^2)$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=-\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1351 : Calculus Ii

Find the derivative of the function:

$f(x)=\cos(e^{2x})+x^5+3x$

Possible Answers:

$-2e^{2x}\sin(e^{2x})+5x^4+3x$

$-2e^{2x}\sin(e^{2x})+5x^4+3$

$2e^{2x}\sin(e^{2x})+5x^4+3$

$2e^{2x}\cos(e^{2x})+5x^4+3$

Correct answer:

$-2e^{2x}\sin(e^{2x})+5x^4+3$

Explanation:

The derivative of the function is equal to

$f'(x)=-2e^{2x}\sin(e^{2x})+5x^4+3$

and was found using the following rules:

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

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Example Question #223 : Derivative Review

Find the second derivative of the function:

$\ln(5x)+\cos(x^2)$

Possible Answers:

$-\frac{5}{x^2}-2\sin(x^2)+4x^2\cos(x^2)$

$\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

$-\frac{5}{x^2}-\sin(x^2)-4x^2\cos(x^2)$

$-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

Correct answer:

$-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{5}{x}-2x\sin(x^2)$

and was found using the following rules:

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

The second derivative of the function is equal to

$f''(x)=-\frac{5}{x^2}-2\sin(x^2)-4x^2\cos(x^2)$

and was found using the same rules as above, along with

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

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Example Question #131 : Derivatives

Find the derivative of the function

$f(x)= \frac{x^2}{x\ln(x)}$

Possible Answers:

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

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

$\frac{\ln(x)-1}{[\ln(x)]^2}$

$\frac{1-\ln(x)}{[\ln(x)]^2}$

None of the other answers.

Correct answer:

$\frac{\ln(x)-1}{[\ln(x)]^2}$

Explanation:

We find the answer using the quotient rule

$(\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

(f(x)g(x))' = f(x)g'(x)+g(x)f'(x)

and then simplifying.

$y' = \frac{x\ln(x)\times 2x-x^2\times(x\frac{1}{x}+\ln(x)))}{(x\ln(x))^2}$

$= \frac{2x^2\ln(x)-x^2-x^2\ln(x)}{x^2\ln(x)^2}$

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

$=\frac{\ln(x)-1}{\ln(x)^2}$ or $\frac{\ln(x)-1}{[\ln(x)]^2}$ . The extra brackets in the denominator are optional.

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Example Question #31 : First And Second Derivatives Of Functions

Find the derivative of the following function at t=0 :

$f(t)=\frac{4e^{\cos(t)}}{50t^2-7t+1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

To find the derivative, we must use the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Now, using the above rule, write out the derivative:

$f'(t)=\frac{(50t^2-7t+1)(-4\sin(t)e^{\cos(t)})-4e^{\cos(t)}(100t-7)}{(50t^2-7t+1)^2}$

The internal derivatives were found using the following rules:

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

Evaluated at t=0 , we get

$f'(0)=\frac{-4(-7)}{1}=28$

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Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function $f(x) = 3xe^{\sin(x)}$ .

Possible Answers:

$3\cos(x)e^{\sin(x)}$

None of the other answers

$3\sin(x)e^{sin(x)}(x\sin(x)+1)$

$3e^{sin(x)}(x\cos(x)+1)$

$3xe^{sin(x)}(x^2\cos(x)+1)$

Correct answer:

$3e^{sin(x)}(x\cos(x)+1)$

Explanation:

We will need to use the product rule and the chain rule to find the derivative.

$y = 3xe^{\sin(x)}$ . Start

$y' = (3x)(e^{\sin(x)})' + (3x)'(e^{\sin(x)})$ . Product Rule

$y' = (3x)(\cos(x)e^{\sin(x)}) + (3)(e^{\sin(x)})$ . Use the Chain Rule for $(e^{\sin(x)})'$ .

$y' = 3x\cos(x)e^{\sin(x)} + 3e^{\sin(x)}$ . Multiply

$y' = 3e^{\sin(x)}(x\cos(x)+1)$ . Factor out a , and an $e^{\sin(x)}$ .

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Example Question #231 : Derivative Review

Find the derivative of the function $y = 50\cos(x)\tan(x)$

Possible Answers:

None of the other answers.

$50\cos^2(x)-\sin^2(x)$

$50\cos(x)sec^{2}(x)$

$50\cos(x)$

$25\sin(x)$

Correct answer:

$50\cos(x)$

Explanation:

Although we could use the Product Rule to compute the derivative, it becomes much easier to find if we rewrite $\tan(x) = \frac{\sin(x)}{\cos(x)}$ .

$y=50\cos(x)\tan(x)$ . Start

$y=50\cos(x)\frac{\sin(x)}{\cos(x)}$

$y = 50\sin(x)$

$y' = 50\cos(x)$ .

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Example Question #31 : First And Second Derivatives Of Functions

If x^y = xy , find in terms of and .

Possible Answers:

$\frac{xy\ln(y)^2}{\ln(x)-xy\ln(x)^2}$

$\frac{xy^2\ln(y)^2}{x\ln(x)-x^2\ln(x)^2}$

None of the other answers

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

$\frac{xy\ln(y)^2}{\ln(x)-x^2y\ln(x)^2}$

Correct answer:

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

Explanation:

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

x^y=yx

$\ln(x^y)=\ln(xy)$

$y\ln(x)=\ln(x)+\ln(y)$

$y = \frac{\ln(y)}{\ln(x)}+1$

$y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}$ . Quotient Rule + implicit differentiation.

$y' = \frac{y'}{y\ln(x)}-\frac{\ln(y)}{x\ln(x)^2}$

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

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

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

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

$y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

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Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function $y= \frac{\tan^{-1}(x)}{x}$

Possible Answers:

$\frac{x^2}{\tan(x)\sqrt{1-x^2}}$

$\frac{x^2}{\tan^{-1}(x)\sqrt{1-x^2}}$

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

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

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}$ .

Using the Quotient Rule and the fact $\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$ , we have

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

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

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

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

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

$y' = \frac{x-\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}$ .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Velocity, Speed, Acceleration

Example Question #21 : First And Second Derivatives Of Functions

Example Question #1351 : Calculus Ii

Example Question #223 : Derivative Review

Example Question #131 : Derivatives

Example Question #31 : First And Second Derivatives Of Functions

Example Question #32 : First And Second Derivatives Of Functions

Example Question #231 : Derivative Review

Example Question #31 : First And Second Derivatives Of Functions

Example Question #32 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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