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Calculus 1 : Differential Functions

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #703 : How To Find Differential Functions

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

Possible Answers:

Answer not listed.

f'(x)= 4x-11

f'(x)= 4x+11

f'(x)= 4x+11x

f'(x)= 4x^2+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

Report an Error

Example Question #703 : How To Find Differential Functions

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

Possible Answers:

Answer not listed.

f'(x)= 4x-11

f'(x)= 4x+11

f'(x)= 4x+11x

f'(x)= 4x^2+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

Report an Error

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

Report an Error

Example Question #711 : How To Find Differential Functions

Find the derivative: $f(x)= e^{3x^2+2x+1}$

Possible Answers:

$f'(x)= e^{6x + 1}$

$f'(x)= e^{3x^2+2x+1}$

$f'(x)= e^{3x^2+2x+1} (6x + 2)$

$f'(x)= e^{3x^2+2x+1} (e^x)$

Answer not listed

Correct answer:

$f'(x)= e^{3x^2+2x+1} (6x + 2)$

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: $f(x)= e^{3x^2+2x+1}$

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

Therefore, the answer is: $f'(x)= e^{3x^2+2x+1} (6x + 2)$

Report an Error

Example Question #711 : How To Find Differential Functions

Find the derivative: $f(x)= e^{3x^2+2x+1}$

Possible Answers:

$f'(x)= e^{6x + 1}$

$f'(x)= e^{3x^2+2x+1}$

$f'(x)= e^{3x^2+2x+1} (6x + 2)$

$f'(x)= e^{3x^2+2x+1} (e^x)$

Answer not listed

Correct answer:

$f'(x)= e^{3x^2+2x+1} (6x + 2)$

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: $f(x)= e^{3x^2+2x+1}$

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

Therefore, the answer is: $f'(x)= e^{3x^2+2x+1} (6x + 2)$

Report an Error

Example Question #711 : Other Differential Functions

Find the first derivative of 2x^3-6x+2 .

Possible Answers:

2x^2-4

2x^2-6

6x^2-4

6x^2-6

None of the other answers

Correct answer:

6x^2-6

Explanation:

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align*} (2x^3-6x+2)' &= 6x^2-6 \end{align*}$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

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 #703 : How To Find Differential Functions

Example Question #703 : How To Find Differential Functions

Example Question #1924 : Calculus

Example Question #711 : How To Find Differential Functions

Example Question #711 : How To Find Differential Functions

Example Question #711 : Other Differential Functions

All Calculus 1 Resources

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