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Calculus 1 : How to find differential functions

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Example Questions

Example Question #2015 : Calculus

Find the derivative:

$f(x)= e^{3x^6 - e^x}$

Possible Answers:

Answer not listed

$f'(x)= e^{x} (18x^{5} - e^x)$

$f'(x)= e^{18x^{5} - e^x}$

$f'(x)= e^{3x^6 - e^x} (18x^{5} - e^x)$

$f'(x)= e^{3x^6 - e^x} (18x^{5} )$

Correct answer:

$f'(x)= e^{3x^6 - e^x} (18x^{5} - e^x)$

Explanation:

If f(x)=C (constant)) , then the derivative is f'(x) = 0 .

If f(x)= x^C , then the derivative is $f'(x) = Cx^{C-1}$ .

If f(x)= ln(x) , then the derivative is $f'(x)= \frac{1}{x}$ .

If f(x)= e^x , the the derivative is f'(x)=e^x .

If $f(x)= \sin x$ , then the derivative is $f'(x)= \cos x$ .

There are many other rules for the derivatives for trig functions.

If F(x)= (f*g)(x) , then the derivative is F'(x)= f'(g(x))g'(x) . This is known as the chain rule.

In this case, we must find the derivative of the following: $f(x)= e^{3x^6 - e^x}$

That is done by doing the following: $f'(x)= e^{3x^6 - e^x} ((6)3x^{6-1} - e^x)$

Therefore, the answer is: $f'(x)= e^{3x^6 - e^x} (18x^{5} - e^x)$

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Example Question #2017 : Calculus

Find the derivative:

$f(x)= e^{ln(x)}$

Possible Answers:

f'(x)= e^x ln(x)

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

Answer not listed

f'(x)= 1

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

Correct answer:

f'(x)= 1

Explanation:

If f(x)=C (constant)) , then the derivative is f'(x) = 0 .

If f(x)= x^C , then the derivative is $f'(x) = Cx^{C-1}$ .

If f(x)= ln(x) , then the derivative is $f'(x)= \frac{1}{x}$ .

If f(x)= e^x , the the derivative is f'(x)=e^x .

If $f(x)= \sin x$ , then the derivative is $f'(x)= \cos x$ .

There are many other rules for the derivatives for trig functions.

If F(x)= (f*g)(x) , then the derivative is F'(x)= f'(g(x))g'(x) . This is known as the chain rule.

In this case, we must find the derivative of the following: $f(x)= e^{ln(x)}$

This simplifies to: $f(x)= e^{ln(x)} =x$

That is done by doing the following: f'(x)= 1

Therefore, the answer is: f'(x)= 1

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Example Question #2016 : Calculus

Determine the slope of the line that is tangent to the function f(x)=x^3+19x+2 at the point x=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.

Taking the derivative of the function

f(x)=x^3+19x+2

The slope of the tangent is

f'(x)=3x^2+19

f'(2)=3(2)^2+19=31

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Example Question #2012 : Calculus

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

Possible Answers:

$\pi$

$-\pi$

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

Product rule: d[uv]=udv+vdu

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

Taking the derivative of the function

f(x)=sin(x)sin(2x)

The slope of the tangent is

f'(x)=cos(x)sin(2x)+sin(x)cos(2x)

$f'(\pi)=cos(\pi)sin(2\pi)+sin(\pi)cos(2\pi)=0$

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Example Question #992 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=15xtan(x) at the point $x=\pi$

Possible Answers:

$-15\pi$

$15\pi$

Correct answer:

$15\pi$

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:

$d[tan(u)]=\frac{du}{cos^2(u)}$

Product rule: d[uv]=udv+vdu

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

Taking the derivative of the function

f(x)=15xtan(x)

The slope of the tangent is

$f'(x)=15tan(x)+\frac{15x}{cos^2(x)}$

$f'(\pi)=15tan(\pi)+\frac{15\pi}{cos^2(\pi)}=15\pi$

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Example Question #2021 : Calculus

Determine the slope of the line that is tangent to the function $f(x)=e^{4x^2}$ at the point x=0.5

Possible Answers:

8e^2

4e^4

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 an exponential:

d[e^u]=e^udu

Taking the derivative of the function

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

The slope of the tangent is

$f'(x)=8xe^{4x^2}$

$f'(0.5)=8(0.5)e^{4(0.5)^2}=4e$

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Example Question #801 : How To Find Differential Functions

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

Possible Answers:

72900

24300

8100

656100

218700

Correct answer:

218700

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.

Taking the derivative of the function

f(x)=3(x^3+3)^3

The slope of the tangent is

f'(x)=27x^2(x^3+3)^2

f'(3)=27(3)^2(3^3+3)^2=218700

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Example Question #991 : Functions

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

Possible Answers:

$2\pi+3$

$2\pi$

$2\pi+2$

Correct answer:

$2\pi+3$

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

The slope of the tangent is

f'(x)=2x+3-sin(x)

$f'(\pi)=2\pi+3-sin(\pi)=2\pi+3$

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Example Question #996 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=\frac{cos(x)}{cos(3x)}$ at the point x=0

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

Quotient rule: $d[\frac{u}{v}]=\frac{vdu-udv}{v^2}$

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

Taking the derivative of the function

$f(x)=\frac{cos(x)}{cos(3x)}$

The slope of the tangent is

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

$f'(0)=\frac{-sin(0)cos(3(0))+3cos(0)sin(3(0))}{cos^2(3(0))}=0$

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Example Question #997 : Differential Functions

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

Possible Answers:

$-2\sqrt{2}$

$4\sqrt{2}$

$-4\sqrt{2}$

Correct answer:

$-4\sqrt{2}$

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

Quotient rule: $d[\frac{u}{v}]=\frac{vdu-udv}{v^2}$

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

Taking the derivative of the function

$f(x)=\frac{sin(4x)}{sin(x)}$

The slope of the tangent is

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

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

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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 #2015 : Calculus

Example Question #2017 : Calculus

Example Question #2016 : Calculus

Example Question #2012 : Calculus

Example Question #992 : Differential Functions

Example Question #2021 : Calculus

Example Question #801 : How To Find Differential Functions

Example Question #991 : Functions

Example Question #996 : Differential Functions

Example Question #997 : Differential Functions

All Calculus 1 Resources

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