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Calculus 1 : How to find differential functions

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Example Questions

Example Question #701 : How To Find Differential Functions

Let f(x)=ln(tan(x)) on the interval $(\frac{\pi}{4},\frac{\pi}{3})$ . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

0.633

-2.509

0.938

0.821

-2.204

Correct answer:

0.938

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint (a,b);a<b , there exists at a minimum one point, , within the interval (a,b) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of f(x)=ln(tan(x)) on the interval $(\frac{\pi}{4},\frac{\pi}{3})$

$f(\frac{\pi}{4})=ln(tan(\frac{\pi}{4}))=0$

$f(\frac{\pi}{3})=ln(tan(\frac{\pi}{3}))=0.549$

Then take the difference of the two and divide by the interval.

$\frac{0.549-0}{\frac{\pi}{3}-\frac{\pi}{4}}=2.097$

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of a natural log:

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

Trigonometric derivative:

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

$f'(x)=\frac{1}{cos^2(x)tan(x)}=\frac{1}{cos(x)sin(x)}$

$\frac{1}{cos(x)sin(x)}=2.097$

Using a calculator, we find the solution x=0.938 , which fits within the interval $(\frac{\pi}{4},\frac{\pi}{3})$ , satisfying the mean value theorem.

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Example Question #701 : How To Find Differential Functions

Find the derivative.

x^3-3x^2+2x

Possible Answers:

3x-6

3x^2-6x+2

x^2-3x+2

Correct answer:

3x^2-6x+2

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}x^3=3x^2$

$\frac{d}{dx}-3x^2=-6x$

$\frac{d}{dx}2x=2$

Thus, the answer is 3x^2-6x+2

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Example Question #702 : How To Find Differential Functions

Find the derivative.

x^3-3x^2+2x

Possible Answers:

3x^2-6x+2

x^2-3x+2

3x-6

Correct answer:

3x^2-6x+2

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}x^3=3x^2$

$\frac{d}{dx}-3x^2=-6x$

$\frac{d}{dx}2x=2$

Thus, the answer is 3x^2-6x+2

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Example Question #704 : How To Find Differential Functions

Find the derivative: $f(x)= 2x^2 + \tan x - (2x-4)^2$

Possible Answers:

Answer not listed

$f'(x)= \sec x - 4x+16$

$f'(x)= \sec^2x$

$f'(x)= \sec^2x - 4x+16$

$f'(x)= \sec^2x - 4x^2+16$

Correct answer:

$f'(x)= \sec^2x - 4x+16$

Explanation:

If f(x)= C (constant) , 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)= e^x , then 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: $f(x)= 2x^2 + \tan x - (2x-4)^2$

That is done by doing the following: $f'(x)= (2)2x^{2-1} + \sec^2x - (2)(2x-4)^{2-1} ( (1) 2x^{1-1} - 0 )$

Therefore, the answer is: $f'(x)= \sec^2x - 4x+16$

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Example Question #705 : How To Find Differential Functions

Find the derivative: $f(x)= \cos(\frac{1}{4}x^4 +3x)$

Possible Answers:

$f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

$f'(x)= \sin(4x^4 +3x) (x^{3} +3)$

$f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

$f'(x)= \cos(\frac{1}{4}x^4 +3x) (x^{3} +3)$

$f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

Answer not listed

$f'(x)= \cos(\frac{1}{4}x^4 +3x) (x^{3} +3)$

Correct answer:

$f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

Explanation:

If f(x)= C (constant) , 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)= e^x , then 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: $f(x)= \cos(\frac{1}{4}x^4 +3x)$

That is done by doing the following: $f'(x)= -\sin(\frac{1}{4}x^4 +3x) ((4)\frac{1}{4}x^{4-1} +(1)3x^{1-1})$

Therefore, the answer is: $f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

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Example Question #1921 : Calculus

Find the derivative: f(x)= (ln(2x))^2

Possible Answers:

Answer not listed

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

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

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

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

Correct answer:

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

Explanation:

If f(x)= C (constant) , 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)= e^x , then 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))^2

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

Therefore, the answer is: $f'(x)= ln(2x) (\frac{2}{x})$

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Example Question #1921 : Calculus

What is the derivative of f(x)=(2x-1)(3x^2+2) ?

Possible Answers:

$f{}'(x)=18x^2-6x$

$f{}'(x)=18x^2-6x+4$

$f{}'(x)=8x^2-6x+4$

$f{}'(x)=18x^2-x+4$

$f{}'(x)=18x^2-6x-4$

Correct answer:

$f{}'(x)=18x^2-6x+4$

Explanation:

Instead of using FOIL to get a polynomial, we can use a special derivative rule, where we multiply the derivative of expression 1 by expression 2 and then add it to the product of teh derivative of expression 2 by expression 1: 2(3x^2+2)+6x(2x-1) . Simplify to get your answer of: $f{}'(x)=18x^2-6x+4$ .

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Example Question #1923 : Calculus

Find the derivative: f(x)= 2(x+2)^2 + 3x

Possible Answers:

f'(x)= 4x+11

f'(x)= 4x^2+11

Answer not listed.

f'(x)= 4x+11x

f'(x)= 4x-11

Correct answer:

f'(x)= 4x+11

Explanation:

If f(x)= C (constant) , 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)= e^x , then 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)= 2(x+2)^2 + 3x

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

Therefore, the answer is: f'(x)= 4x+11

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Example Question #1923 : Calculus

Find the derivative: f(x)= 2(x+2)^2 + 3x

Possible Answers:

f'(x)= 4x+11

f'(x)= 4x^2+11

Answer not listed.

f'(x)= 4x+11x

f'(x)= 4x-11

Correct answer:

f'(x)= 4x+11

Explanation:

If f(x)= C (constant) , 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)= e^x , then 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)= 2(x+2)^2 + 3x

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

Therefore, the answer is: f'(x)= 4x+11

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Example Question #1924 : Calculus

Find the derivative: f(x)= 5 ln (5x^2-3x+1)

Possible Answers:

$f'(x)=\frac{10x+ 3}{5x^2+3x-1}$

Answer not listed

$f'(x)=ln\frac{50x+ 15}{5x^2+3x-1}$

$f'(x)=\frac{10x+ 3}{5x^2+3x-1}$

Answer not listed

$f'(x)=\frac{50x+ 15}{5x^2+3x-1}$

$f'(x)=\frac{5}{5x^2+3x-1}$

$f'(x)=ln\frac{50x+ 15}{5x^2+3x-1}$

$f'(x)=\frac{5}{5x^2+3x-1}$

Correct answer:

$f'(x)=\frac{50x+ 15}{5x^2+3x-1}$

Explanation:

If f(x)= C (constant) , 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)= e^x , then 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)= 5 ln (5x^2-3x+1)

That is done by doing the following: $f'(x)=( 5) \frac{1}{5x^2+3x-1} ((2)5x^{2-1} + (1)3x^{1-1} - 0)$

Therefore, the answer is: $f'(x)=\frac{50x+ 15}{5x^2+3x-1}$

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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 #701 : How To Find Differential Functions

Example Question #701 : How To Find Differential Functions

Example Question #702 : How To Find Differential Functions

Example Question #704 : How To Find Differential Functions

Example Question #705 : How To Find Differential Functions

Example Question #1921 : Calculus

Example Question #1921 : Calculus

Example Question #1923 : Calculus

Example Question #1923 : Calculus

Example Question #1924 : Calculus

All Calculus 1 Resources

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