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Calculus 2 : First and Second Derivatives of Functions

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Example Questions

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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Example Question #1361 : Calculus Ii

Find the derivative of the function $y=e^{x^{\frac{1}{2}}+x^2-\ln(x)}$

Possible Answers:

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

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

None of the other answers

$\frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x}$

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

Correct answer:

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

Explanation:

The Chain Rule is required here.

$y=e^{x^{\frac{1}{2}}+x^2-\ln(x)}$ . Start

The Chain Rule Proceeds as follows: $f(g(x)) = f'(g(x))\times g'(x)$ . In this case

f(x)= e^x

and

$g(x) = \frac{1}{2}x^{-\frac{1}{2}}+2x-\frac{1}{x}$ .

Putting these into the Chain Rule, we get

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

Or the same to say

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

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Example Question #231 : Derivative Review

Evaluate the derivative of y = 6x^2 +C , where is any constant.

Possible Answers:

12x+1

6x^2 +C

12x

None of the other answers

Correct answer:

12x

Explanation:

For the 6x^2 term, we simply use the power rule to abtain 12x . Since is a constant (not a variable), we treat it as such. The derivative of any constant (or "stand-alone number") is .

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Example Question #32 : First And Second Derivatives Of Functions

Find the derivative of the function y=7x^2e^x .

Possible Answers:

14x^2(x-1)

$\frac{7}{3}x^3e^x$

14xe^x

7xe^x(x+2)

None of the other answers

Correct answer:

7xe^x(x+2)

Explanation:

We use the Product Rule to find our answer here. The Product Rule formula is (fg)'(x) = f(x)g'(x)+f'(x)g(x) .

Let f(x) = 7x^2 , g(x)= e^x , then we have f'(x) = 14x , g'(x)=e^x .

Putting these into our formula, we have

y' = (7x^2)(e^x)+(e^x)(14x)

=e^x(7x^2+14x)

=7xe^x(x+2) .

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Example Question #1364 : Calculus Ii

What is the derivative of

$f(x)=5\sqrt{x}$ ?

Possible Answers:

$f'(x)=\frac{5}{2\sqrt{x}}$

$f'(x)=\frac{5}{2}\sqrt{x}$

$f'(x)=\frac{5}{\sqrt{x}}$

$f'(x)=\frac{2}{\sqrt{x}}$

Correct answer:

$f'(x)=\frac{5}{2\sqrt{x}}$

Explanation:

We can find the derivative of

$f(x)=5\sqrt{x}$

using the power rule

$\frac{d}{dx}x^a=ax^{a-1}$

with $a=\frac{1}{2}$

so we have

$f'(x)=5\frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{5}{2x^{1/2}}=\frac{5}{2\sqrt{x}}$

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Example Question #1365 : Calculus Ii

Find the velocity function given the displacement function:

$f(t)=\pi t^{2/5}$

Possible Answers:

$f'(t)=\frac{\pi}{ t^{3/5}}$

$f'(t)=\frac{2\pi}{ 5t^{2/5}}$

$f'(t)=\frac{2\pi}{ 5t^{3/5}}$

$f'(t)=\frac{5\pi}{ 2t^{3/5}}$

Correct answer:

$f'(t)=\frac{2\pi}{ 5t^{3/5}}$

Explanation:

The derivative of the displacement function is the velocity, so we need to find f'(t) . We can use the power rule

$\frac{d}{dx} x^a=ax^{a-1}$

with $a=\frac{2}{5}$

to get

$f'(t)=\pi\frac{2}{ 5}t^{-3/5}=\frac{2\pi}{ 5t^{3/5}}$

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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #1361 : Calculus Ii

Example Question #231 : Derivative Review

Example Question #32 : First And Second Derivatives Of Functions

Example Question #1364 : Calculus Ii

Example Question #1365 : Calculus Ii

All Calculus 2 Resources

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