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Calculus 1 : How to find differential functions

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Example Questions

Example Question #2051 : Calculus

Determine the slope of the line that is tangent to the function f(x)=cos(4x) at the point $x=\frac{\pi}{2}$

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

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

Taking the derivative of the function f(x)=cos(4x) at the point $x=\frac{\pi}{2}$

The slope of the tangent is

f'(x)=-4sin(4x)

$f'(\frac{\pi}{2})=-4sin(4(\frac{\pi}{2}))=0$

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Example Question #1026 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=cos^2(\pi x^2)$ at the point $x=\frac{1}{2}$

Possible Answers:

$-\pi$

$-4\pi$

$\pi$

$-2\pi$

Correct answer:

$-\pi$

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

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

Taking the derivative of the function $f(x)=cos^2(\pi x^2)$ at the point $x=\frac{1}{2}$

The slope of the tangent is

$f'(x)=-4\pi x sin(\pi x^2)cos(\pi x^2)$

$f'(\frac{1}{2})=-4\pi (\frac{1}{2}) sin(\pi (\frac{1}{2})^2)cos(\pi (\frac{1}{2})^2)=-\pi$

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Example Question #1027 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=cos(sin(x)) at the point $x=3\pi$

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

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

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

Taking the derivative of the function f(x)=cos(sin(x)) at the point $x=3\pi$

The slope of the tangent is

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

$f'(3\pi)=cos(3\pi)sin(sin(3\pi))=0$

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Example Question #841 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=ln(x^6) at the point x=1

Possible Answers:

$\frac{1}{6}$

$\frac{5}{6}$

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):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

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

Taking the derivative of the function f(x)=ln(x^6) at the point x=1

The slope of the tangent is

$f'(x)=\frac{6x^5}{x^6}=\frac{6}{x}$

$f'(1)=\frac{6}{1}=6$

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Example Question #842 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=ln(2x^4) at the point x=2

Possible Answers:

ln(64)

ln(2)

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):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

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

Taking the derivative of the function

The slope of the tangent is f(x)=ln(2x^4) at the point x=2

$f'(x)=\frac{8x^3}{2x^4}=\frac{4}{x}$

$f'(2)=\frac{4}{2}=2$

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Example Question #1032 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=ln(x^2+x) at the point x=3

Possible Answers:

$\frac{7}{12}$

$\frac{1}{12}$

ln(7)

$\frac{1}{7}$

Correct answer:

$\frac{7}{12}$

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):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

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

Taking the derivative of the function f(x)=ln(x^2+x) at the point x=3

The slope of the tangent is

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

$f'(3)=\frac{2(3)+1}{3^2+3}=\frac{7}{12}$

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Example Question #843 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=e^{x^3}$ at the point x=2

Possible Answers:

$e^{12}$

12e^8

8e^8

$e^{8}$

$12e^{12}$

Correct answer:

12e^8

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):

Derivative of an exponential:

d[e^u]=e^udu

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

Taking the derivative of the function $f(x)=e^{x^3}$ at the point x=2

The slope of the tangent is

$f'(x)=3x^2e^{x^3}$

$f'(2)=3(2)^2e^{2^3}=12e^8$

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Example Question #1033 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=5e^{5x}$ at the point x=3

Possible Answers:

$25e^{5}$

$25e^{15}$

$15e^{15}$

5e^5

$5e^{15}$

Correct answer:

$25e^{15}$

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):

Derivative of an exponential:

d[e^u]=e^udu

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

Taking the derivative of the function $f(x)=5e^{5x}$ at the point x=3

The slope of the tangent is

$f'(x)=25e^{5x}$

$f'(3)=25e^{5(3)}=25e^{15}$

Report an Error

Example Question #1034 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=e^{2x+4}$ at the point x=4

Possible Answers:

$2e^{12}$

$e^{12}$

$12e^{2}$

$12e^{12}$

$6e^{12}$

Correct answer:

$2e^{12}$

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):

Derivative of an exponential:

d[e^u]=e^udu

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

Taking the derivative of the function $f(x)=e^{2x+4}$ at the point x=4

The slope of the tangent is

$f'(x)=2e^{2x+4}$

$f'(4)=2e^{2(4)+4}=2e^{12}$

Report an Error

Example Question #1031 : Functions

Find the first derivative of the following function.

f(x)=2x^4-3x^3+2x^2-18x+5

Possible Answers:

f'(x)=8x^5-9x^4+4x^3-18x^2+5x

None of the other answers.

f'(x)=8x^3-9x^2+4x-18

f'(x)=16x^3-27x^2+4x-18

f'(x)=2x^3-3x^2+2x-18

Correct answer:

f'(x)=8x^3-9x^2+4x-18

Explanation:

The first derivative of the given function makes use of the power rule of derivatives.

If you have a function

f(x)=ax^n

According to the Power Rule, the first derivative is defined as

f'(x)=nax^n^-^1

If this rule is applied to each of the terms in the function f(x) we get that

f'(x)=4(2)x^4^-^1-3(3)x^3^-^1+2(2)x^2^-^1-1(18)x^1^-^1+0(5)x^0^-^1

=8x^3-9x^2+4x-18

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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 #2051 : Calculus

Example Question #1026 : Differential Functions

Example Question #1027 : Differential Functions

Example Question #841 : How To Find Differential Functions

Example Question #842 : How To Find Differential Functions

Example Question #1032 : Differential Functions

Example Question #843 : How To Find Differential Functions

Example Question #1033 : Differential Functions

Example Question #1034 : Differential Functions

Example Question #1031 : Functions

All Calculus 1 Resources

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