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Calculus 1 : How to find differential functions

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Example Questions

Example Question #661 : Other 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:

3.0

$\pi$

$-\pi$

-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 #844 : 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:

$\frac{\pi}{4}$

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

$\frac{\pi}{2}$

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

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 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 #845 : 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)

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

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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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Example Question #661 : How To Find Differential Functions

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

Possible Answers:

$4\pi$

$-4\pi$

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):

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(4x^2)) at the point $x=\frac{\pi}{2}$

The slope of the tangent is

$f'(x)=-\frac{-8xsin(4x^2)}{cos(4x^2)}=8xtan(4x^2)$

$f'(\frac{\pi}{2})=8(\frac{\pi}{2})tan(4(\frac{\pi}{2})^2)=5.99 \approx ]6.0$

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Example Question #663 : How To Find Differential Functions

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

Possible Answers:

2.89

$\frac{\pi}{4}$

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

-2.89

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

The slope of the tangent is

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

$f'(-\pi)=tan(-\pi)sin(ln(cos(-\pi)))=0$

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Example Question #851 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=(x^3+2x+4)^3 at the point x=1 .

Possible Answers:

735

5145

147

245

Correct answer:

735

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+2x+4)^3 at the point x=1

The slope of the tangent is

f'(x)=3(3x^2+2)(x^3+2x+4)^2

f'(1)=3(3(1)^2+2)(1^3+2(1)+4)^2=735

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Example Question #665 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=(2x^3-8x+5)^5 at the point x=1 .

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)=(2x^3-8x+5)^5 at the point x=1

The slope of the tangent is

f'(x)=5(6x^2-8)(2x^3-8x+5)^4

f'(1)=5(6(1)^2-8)(2(1)^3-8(1)+5)^4=-10

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Example Question #669 : Other Differential Functions

Find the derivative: $f(x) = 12x^{\frac{1}{2}} + (6x^2 - 5x +2)^2$

Possible Answers:

$f'(x) = 6x^{\frac{1}{2}} + (6x^2 - 5x +2) (12x- 5 )$

$f'(x) = 6x^{-\frac{1}{2}} - 2(6x^2 - 5x +2) (12x- 5x)$

$f'(x) = 6x^{-\frac{1}{2}} + 2(6x^2 - 5x +2) (12x- 5 )$

$f'(x) = 6x^{-\frac{1}{2}} + 2(6x^2 - 5x +2)$

f'(x) = 2(6x^2 - 5x +2) (12x- 5 )

Correct answer:

$f'(x) = 6x^{-\frac{1}{2}} + 2(6x^2 - 5x +2) (12x- 5 )$

Explanation:

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

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

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

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

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

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: $f(x) = 12x^{\frac{1}{2}} + (6x^2 - 5x +2)^2$

That is done by doing the following: $f'(x) = (\frac{1}{2})12x^{(\frac{1}{2}-1)} + (2)(6x^2 - 5x +2)^{(2-1)}((2)6x^{(2-1)}- (1)5x^{(1-1)}+0 )$

Therefore, the answer is: $f'(x) = 6x^{-\frac{1}{2}} + 2(6x^2 - 5x +2) (12x- 5 )$

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Example Question #661 : How To Find Differential Functions

Find the derivative: $f(x) = \sin(x^2+2x-3)$

Possible Answers:

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

$f'(x) = -\cos (x^2+2x-3) (2x + 2)$

$f'(x) = \cos (x^2+2x-3) (2x + 2)^2$

$f'(x) = \cos (x^2+2x-3) (2x + 2)$

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

Correct answer:

$f'(x) = \cos (x^2+2x-3) (2x + 2)$

Explanation:

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

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

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

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

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

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: $f(x) = \sin(x^2+2x-3)$

That is done by doing the following: $f'(x) = \cos (x^2+2x-3) ((2)x^{(2-1)} + (1)2x^{(1-1)} + 0)$

Therefore, the answer is: $f'(x) = \cos (x^2+2x-3) (2x + 2)$

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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 #661 : Other Differential Functions

Example Question #844 : Functions

Example Question #845 : Functions

Example Question #661 : How To Find Differential Functions

Example Question #661 : How To Find Differential Functions

Example Question #663 : How To Find Differential Functions

Example Question #851 : Differential Functions

Example Question #665 : How To Find Differential Functions

Example Question #669 : Other Differential Functions

Example Question #661 : How To Find Differential Functions

All Calculus 1 Resources

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