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Calculus 1 : How to find differential functions

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Example Questions

Example Question #2047 : Calculus

Determine if the function is differentiable for all :

$f(t)=\begin{Bmatrix} \frac{1}{x}, x<-1\\ ln(-x-1+e), x\geq -1 \end{Bmatrix}$

Possible Answers:

The function is not differentiable for all

The answer cannot be determined without analysis in the complex plane

The function is differentiable for all

The function is differentiable but not continuous for all

Correct answer:

The function is not differentiable for all

Explanation:

When looking at differentiability of piecewise functions over all , first consider if the two functions are continuous and differentiable for all x.

$\frac{1}{x}$ is discontinuous at x=0 , but that's okay $\frac{1}{x}$ is restricted for x<-1 . However ln(-x) does not exist for $x\geq0$ . Since this is the case, we can say that this piecewise function is not differentiable for all .

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Example Question #2048 : Calculus

What is the second derivative of f(x)=6x^3-2x^2+9 ?

Possible Answers:

f{}''(x)=36x

f{}''(x)=36x-4

f{}''(x)=4x

f{}''(x)=-36x-4

f{}''(x)=36x+4

Correct answer:

f{}''(x)=36x-4

Explanation:

To find the second derivative, you must first find the first derivative. When taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent. Therefore, the first derivative is: 18x^2-4x . Then, take the derivative again to get the second derivative: f{}''(x)=36x-4 .

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Example Question #831 : How To Find Differential Functions

What is the derivative of f(x)=(4x^2-1)(9)

Possible Answers:

f{}'(x)=72x

f{}'(x)=2x

f{}'(x)=72x+1

f{}'(x)=36x

f{}'(x)=-72x

Correct answer:

f{}'(x)=72x

Explanation:

First, distribute the 9 to get 36x^2-9 . Then, take the derivative, remembering to multiply the exponent by the coefficient in front of the x and then subtracting one from the exponent to get an answer of f{}'(x)=72x .

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Example Question #1021 : Functions

Find the first derivative of the following function:

$f(x)=3xe^{x}+12\cos(x)$

Possible Answers:

$3e^x+3xe^x-\sin(x)$

$3e^x+3xe^x-12\sin(x)$

$3e^x+3x^2e^x-12\sin(x)$

$3e^x+3xe^x+12\sin(x)$

$3+3xe^x-12\sin(x)$

Correct answer:

$3e^x+3xe^x-12\sin(x)$

Explanation:

The derivative of the function is equal to

$f'(x)=3e^x+3xe^x-12\sin(x)$

and was found using the following rules:

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

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Example Question #1021 : Differential Functions

Find the derivative of the following function at x=1:

$(f\circ g)(x)$

where f(x)=2x^2 , g(x)=5x

Possible Answers:

100

Correct answer:

100

Explanation:

To find the derivative of the composite function, we must rewrite the composition as f(g(x)) . Using the chain rule, the derivative of this function is equal to

$f'(g(x)) \cdot g'(x)$ .

For our function,

f'(x)=4x and g'(x)=5 . These derivatives were found using the following rule:

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

We now have all the pieces to use the chain rule formula, where we are evaluating the function at x=1:

g(1)=5

g'(1)=5

f'(g(1))=f'(5)=20

$20 \cdot 5=100$

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Example Question #1021 : Differential Functions

Find the first derivative of the function:

$f(x)= \frac{t^4-t^2}{e^t}$

Possible Answers:

e^t(2t-t^2)

$\frac{-t^4+4t^3+t^2-2t}{e^{2t}}$

e^t(-t^4+4t^3+t^2-2t)

e^t(t^4+4t^3+t^2+2t)

Correct answer:

e^t(-t^4+4t^3+t^2-2t)

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{e^t(4t^3-2t)-e^t(t^4+t^2)}{e^{2t}}=e^t(-t^4+4t^3+t^2-2t)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Example Question #1022 : Differential Functions

Find the second derivative of the function:

$f(x)=\cos(\sec(e^{10x^2}))$

Possible Answers:

$20xe^{10x^2}$

$20xe^{10x^2}\sec(x)\tan(x)$

$-20xe^{10x^2}$

Correct answer:

$20xe^{10x^2}$

Explanation:

Because $\cos(x)$ and $\sec(x)$ are inverses of each other, we can simplify the given function to

$f(x)=e^{10x^2}$

Taking the derivative of this, we get

$f'(x)=20xe^{10x^2}$

by using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1023 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=sin(10x) at the point $x=\pi$

Possible Answers:

Correct answer:

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=sin(10x) at the point $x=\pi$

The slope of the tangent is

f'(x)=10cos(x)

$f'(\pi)=10cos(\pi)=-10$

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Example Question #1024 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=sin(5\pi x^2)$ at the point x=2

Possible Answers:

$20\pi$

$10\pi$

$-10\pi$

$-20\pi$

Correct answer:

$20\pi$

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=sin(5\pi x^2)$ at the point x=2

The slope of the tangent is

$f'(x)=10\pi x cos(5\pi x^2)$

$f'(2)=10\pi (2) cos(5\pi (2)^2)=20\pi$

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Example Question #1025 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=sin^2(5x) at the point $x=\pi$

Possible Answers:

Correct answer:

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=sin^2(5x) at the point $x=\pi$

The slope of the tangent is

f'(x)=10sin(5x)cos(5x)

$f'(\pi)=10sin(5\pi)cos(5\pi)=0$

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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 #2047 : Calculus

Example Question #2048 : Calculus

Example Question #831 : How To Find Differential Functions

Example Question #1021 : Functions

Example Question #1021 : Differential Functions

Example Question #1021 : Differential Functions

Example Question #1022 : Differential Functions

Example Question #1023 : Differential Functions

Example Question #1024 : Differential Functions

Example Question #1025 : Differential Functions

All Calculus 1 Resources

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