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Calculus 1 : Calculus

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Example Questions

Example Question #841 : Differential Functions

f(x)= 5x^3 + (2x - ln(x))^2

Find the derivative.

Possible Answers:

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

f'(x)= 15x^2 + 2(2x - ln(x))

$f'(x)= 15x^2 + (2-\frac{1}{x})$

$f'(x)= 15x^2 + 2(2x - ln(x))(2-\frac{1}{x})$

f'(x)= 15x^2

Correct answer:

$f'(x)= 15x^2 + 2(2x - ln(x))(2-\frac{1}{x})$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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)= 5x^3 + (2x - ln(x))^2

That is done by doing the following: $f'(x)= (3)5x^{(3-1)} + (2)(2x - ln(x))^{(2-1)} (2-\frac{1}{x})$

Therefore, the answer is: $f'(x)= 15x^2 + 2(2x - ln(x))(2-\frac{1}{x})$

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Example Question #843 : Functions

$f(x)= ln(2x) + \cos (\sin x)$

Find the derivative.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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: $f(x)= ln(2x) + \cos (\sin x)$

That is done by doing the following: $f'(x)= (\frac{1}{2x}) (2) + (-\sin (\sin x)) (\cos x)$

Therefore, the answer is: $f'(x)= \frac{1}{x} -\sin (\sin x) (\cos x)$

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Example Question #651 : Other Differential Functions

Determine the slope of the line that is tangent to the function f(x)=sin(x^6) at the point x=1.7 . Use degrees for calculations.

Possible Answers:

-31.3

85.9

77.7

-75.2

41.6

Correct answer:

77.7

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(x^6) at the point x=1.7

The slope of the tangent is

f'(x)=6x^5cos(x^6)

f'(1.7)=6(1.7)^5cos(1.7^6)=77.7

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Example Question #841 : Differential Functions

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

Possible Answers:

$-\frac{\pi}{2}$

$\pi$

$\frac{\pi}{2}$

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:

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

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)=-ln(cos(x)) at the point $x=\frac{\pi}{4}$

The slope of the tangent is

$f'(x)=-\frac{-sin(x)}{cos(x)}=tan(x)$

$f'(\frac{\pi}{4})=tan(\frac{\pi}{4})=1$

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Example Question #841 : Differential Functions

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

Possible Answers:

$-\pi$

$-\frac{\pi}{2}$

$\frac{\pi}{2}$

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:

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

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)=ln(-cos(x)) at the point $x=\frac{\pi}{4}$

The slope of the tangent is

$f'(x)=\frac{sin(x)}{-cos(x)}=-tan(x)$

$f'(\frac{\pi}{4})=-tan(\frac{\pi}{4})=-1$

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Example Question #651 : Other Differential Functions

Determine the slope of the line that is tangent to the function f(x)=ln(sin(x)) at the point $x=\frac{\pi}{3}$ .

Possible Answers:

$-\frac{\sqrt{3}}{3}$

$-\sqrt{3}$

$\sqrt{3}$

$\frac{\sqrt{3}}{3}$

Correct answer:

$\frac{\sqrt{3}}{3}$

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):

Derivative of a natural log:

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

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)=ln(sin(x)) at the point $x=\frac{\pi}{3}$

The slope of the tangent is

$f'(x)=\frac{cos(x)}{sin(x)}=cot(x)$

$f'(\frac{\pi}{3})=cot(\frac{\pi}{3})=\frac{\sqrt{3}}{3}$

Report an Error

Example Question #661 : How To Find Differential Functions

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

Possible Answers:

$\pi$

$-\pi$

3.0

-3.0

Correct answer:

3.0

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:

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

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)=-ln(cos(x^2)) at the point $x=\pi$

The slope of the tangent is

$f'(x)=-\frac{-2xsin(x^2)}{cos(x^2)}=2xtan(x^2)$

$f'(\pi)=2\pi tan(\pi^2)=3$

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Example Question #661 : Other Differential Functions

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

Possible Answers:

$\frac{\pi}{4}$

$-\frac{\pi}{4}$

$\frac{\pi}{2}$

$-\frac{\pi}{2}$

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:

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

Trigonometric derivative:

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

Product rule: d[uv]=udv+vdu

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

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

The slope of the tangent is

$f'(x)=ln(sin(x))+\frac{xcos(x)}{sin(x)}=ln(sin(x))+xcot(x)$

$f'(\frac{\pi}{2})=ln(sin(\frac{\pi}{2}))+\frac{\pi}{2}cot(\frac{\pi}{2})=0$

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Example Question #841 : Differential Functions

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

Possible Answers:

$-\frac{5\pi}{2}$

$\frac{5\pi}{2}$

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):

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(x) at the point $x=\frac{\pi}{2}$

The slope of the tangent is

f'(x)=5cos(x)sin^4(x)

$f'(\frac{\pi}{2})=5cos(\frac{\pi}{2})sin^4(\frac{\pi}{2})=0$

Report an Error

Example Question #661 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=x^3tan(4x) at the point $x=\pi$ .

Possible Answers:

$4\pi^3$

$\pi$

$\pi^3$

Correct answer:

$4\pi^3$

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:

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

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)=x^3tan(4x) at the point $x=\pi$

The slope of the tangent is

$f'(x)=3x^2tan(4x)+\frac{4x^3}{cos^2(4x)}$

$f'(\pi)=3\pi^2tan(4\pi)+\frac{4\pi^3}{cos^2(4\pi)}=4\pi^3$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #841 : Differential Functions

Example Question #843 : Functions

Example Question #651 : Other Differential Functions

Example Question #841 : Differential Functions

Example Question #841 : Differential Functions

Example Question #651 : Other Differential Functions

Example Question #661 : How To Find Differential Functions

Example Question #661 : Other Differential Functions

Example Question #841 : Differential Functions

Example Question #661 : How To Find Differential Functions

All Calculus 1 Resources

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