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Calculus 1 : How to find differential functions

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Example Questions

Example Question #373 : Functions

Find the derivative of the function f(x)=cos(x^3+2x^2+x+3) .

Possible Answers:

(3x^2+4x+1)cos(x^3+2x^2+x+3)

-(3x^2+4x+1)sin(3x^2+4x+1)

(3x^2+4x+1)cos(3x^2+4x+1)

-sin(x^3+2x^2+x+3)

-(3x^2+4x+1)sin(x^3+2x^2+x+3)

Correct answer:

-(3x^2+4x+1)sin(x^3+2x^2+x+3)

Explanation:

To find the derivative of this function, f(x)=cos(x^3+2x^2+x+3) , utilize the chain rule:

cos(g(x))'=-g'(x)sin(g(x))

Where

g(x)=x^3+2x^2+x+3

Performing the derivative gives:

g'(x)=3x^2+4x+1

So the derivative is:

f'(x)=-(3x^2+4x+1)sin(x^3+2x^2+x+3)

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Example Question #191 : How To Find Differential Functions

Find the derivative of the function $f(x)=\frac{e^{x^2}}{ln(x)}$ .

Possible Answers:

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

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

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

${2x^2e^{x^2}ln(x)-e^{x^2}}$

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

Correct answer:

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

Explanation:

For the function $f(x)=\frac{e^{x^2}}{ln(x)}$ , it will be useful to utilize the chain rule and quotient rule of derivatives.

The quotient rule states:

$\frac{d}{dx}\frac{u}{v}=\frac{vdu-udv}{v^2}$

Thus

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

The ln(x)' value is simply $\frac{1}{x}$ .

To find the value of $e^{x^2}'$ , note that it should follow the form

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

$e^{x^2}'=2xe^{x^2}$

Putting everything together:

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

$f'(x)=\frac{2x^2e^{x^2}ln(x)-e^{x^2}}{xln^2(x)}$

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Example Question #375 : Functions

What is the first derivative of $f(x)=3x^{2}\cos(x)$ ?

Possible Answers:

$-x\sin(x+1)$

$2\cos(2x)$

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

$6x\cos(x)-4\sin(x)$

$6x\cos(x)-3x^{2}\sin(x)$

Correct answer:

$6x\cos(x)-3x^{2}\sin(x)$

Explanation:

To solve this problem, we must use the following formulae:

$\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x), \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$\frac{d}{dx}\cos(x)=-\sin(x)$

Where $f(x)=3x^{2}$ and $g(x)=\cos(x)$

$\frac{d}{dx}(3x^{2}\cos(x))=3*2x^{2-1}\cos(x)-3x^{2}\sin(x)=6x\cos(x)-3x^{2}\sin(x)$

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Example Question #192 : How To Find Differential Functions

Find the derivative of $f(x)=\frac{ln(x)}{1+sin(x)}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find this derivative, we need to use the following formulae:

$\frac{d}{dx}(\frac{g(x)}{h(x)})=\frac{g'(x)h(x)-g(x)h'(x)}{g(x)^{2}}$

$\frac{d}{dx} \sin(x)= \cos(x), \frac{d}{dx} \ln(x)=\frac{1}{x}, \frac{d}{dx}(c)=0$

Where $g(x)=\ln(x)$ and $h(x)=1+\sin(x)$ .

$\frac{\frac{1}{x}*(1+sin(x))-cos(x)ln(x))}{(sin(x)+1)^{2}}$ $=\frac{\frac{1+sin(x)}{x}-cos(x)ln(x)}{sin^{2}(x)+2sin(x)+1}$

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Example Question #1411 : Calculus

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

Possible Answers:

$\csc(x)\cot(x)\ln(3)*3^{csc(x)}$

$-\csc(x)\cot(x)*3^{\csc(x)}$

$-\csc(x)\cot(x)\ln(3)*3^{\csc(x)}$

$-\cot(x)\ln(3)*3^{\csc(x)}$

$-\csc(x)\ln(3)*3^{csc(x)}$

Correct answer:

$-\csc(x)\cot(x)\ln(3)*3^{\csc(x)}$

Explanation:

To find the derivative of f(x) , we need the following formulae:

$\frac{d}{dx} \csc(x)=-\csc(x)\cot(x), \frac{d}{dx} f(g(x))=f'(g(x))g'(x)$

$\frac{d}{dx} a^{x}=a^{x}\ln(a)$

$f'(x)=\ln(3)*3^{\csc(x)}*-\csc(x)\cot(x)$

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Example Question #1412 : Calculus

Find f'(x) if $f(x)=\log_{4}(x^{3})$ .

Possible Answers:

$\frac{3}{ln(4)}$

$\frac{-3}{xln(4)}$

$\frac{3}{4xln(4)}$

$\frac{3}{xln(4)}$

$\frac{1}{xln(4)}$

Correct answer:

$\frac{3}{xln(4)}$

Explanation:

To find the derivative of f(x), we must use the following formulae:

$\\ \frac{d}{dx}(\log{_{a}}(x))=\frac{1}{x\ln(a)},\\ \frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)$

In this particular case,

$\\a=4, \\ g(x)=x^3$

thus our derivative becomes,

$f'(x)= \frac{1}{x^{3}ln(4)}*3x^{2} = \frac{3}{xln(4)}$ .

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Example Question #1413 : Calculus

Find the first derivative of $f(x)=\ln(2x^{2})$ .

Possible Answers:

$\frac{2}{3x}$

$\frac{2}{x}$

$\frac{2}{x^{2}}$

$\frac{4}{x}$

$\frac{-2}{x}$

Correct answer:

$\frac{2}{x}$

Explanation:

To find the first derivative of f(x) , we must use the following formulae:

$\frac{d}{dx} \ln(x)=\frac{1}{x}, \frac{d}{dx} f(g(x))=f'(g(x))g'(x)$

Applying these rules we can find the following derivative.

Let,

$\\f(x)=\frac{1}{2x^2}\\ \\g(x)=2x^2$

therefore we get,

$f'(x)= \frac{1}{2x^{2}}*4x= \frac{2}x{}$ .

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Example Question #1414 : Calculus

Find the first derivative of $f(x)=x^{4}\cos(2x)$ .

Possible Answers:

$4x^{3}\cos(2x)-2x^{4}\sin(2x)$

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

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

$x^{3}\cos(2x)-x^{4}\sin(2x)$

$4x^{3}\cos(2x)-2x\sin(2x)$

Correct answer:

$4x^{3}\cos(2x)-2x^{4}\sin(2x)$

Explanation:

To find this derivative, we need to use the following formulae:

$\\ \frac{d}{dx}(x^{n}))=nx^{n-1}, \\ \\ \frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

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

Applying these rules where

$\\ f(x)=x^4 \\ \\ g(x)=\cos(2x)$

therefore we get,

$f'(x)=4x^{3}\cos(2x)-x^{4}\sin(2x)*2$ .

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Example Question #1415 : Calculus

Find the derivative of $f(x)=3x^{2}\tan(7x)$ .

Possible Answers:

$x\tan(7x)+21x^{2}\sec^{2}(7x)$

$6x\tan(7x)+21x^{2}\sec^{2}(7x)$

$6x\tan(7x)+3x^{2}\sec^{2}(7x)$

$6x\tan(7x)+21x\sec^{2}(7x)$

$6x\tan(7x)$

Correct answer:

$6x\tan(7x)+21x^{2}\sec^{2}(7x)$

Explanation:

To find this derivative, we must use the product rule, power rule, chain rule, and trigonometric derivative rule for tangent.

Lets recall the product rule,

$\frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

In this particular problem,

f(x)=3x^2 and g(x)=tan(7x) .

In order to find f'(x) we will need to use the power rule which states,

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

Therefore,

$f'(x)=2\cdot 3x^{2-1}=6x$ .

To find g'(x) we need to use the chain rule which states,

$\frac{d}{dx} h(i(x))=h'(i(x))i'(x)$

where h(x)=tan(7x) and i(x)=7x .

To find h'(x) we will need to use the trigonometric derivative rule for tangent which states,

$\frac{d}{dx}tan(x)=sec^2(x)$ and to find i'(x) we will again use the power rule.

Thus,

h'(x)=sec^2(7x) and $i'(x)=7\cdot 1x^{1-1}=7$ .

This then makes,

g'(x)=7sec^2(7x) .

Now lets combine our terms using the product rule to find the final derivative.

$\\f'(x)=6x\tan(7x)+3x^{2}\cdot 7sec^{2}(7x) \\ f'(x)= 6x\tan(7x)+21x^{2}sec^{2}(7x)$

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Example Question #1416 : Calculus

What is the derivative of $x^{e^{-2x}}$ ?

Possible Answers:

$-2e^{-2x}x^{e^{-2x}-1}$

$-2e^{-2x}$

$-2e^{-2x}x^{e^{-2x}}$

$e^{-2x}x^{e^{-2x}-1}$

$-2e^{-2x}x$

Correct answer:

$-2e^{-2x}x^{e^{-2x}-1}$

Explanation:

This problem relies on the chain rule

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$ .

First, you have x to something, so you must use the power rule

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

to get

$(e^{-2x})'x^{e^{-2x}-1}$ .

Using the chain rule, this becomes $-2e^{-2x}x^{e^{-2x}-1}$ .

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #373 : Functions

Example Question #191 : How To Find Differential Functions

Example Question #375 : Functions

Example Question #192 : How To Find Differential Functions

Example Question #1411 : Calculus

Example Question #1412 : Calculus

Example Question #1413 : Calculus

Example Question #1414 : Calculus

Example Question #1415 : Calculus

Example Question #1416 : Calculus

All Calculus 1 Resources

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