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Calculus 1 : Differential Functions

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Example Questions

Example Question #794 : How To Find Differential Functions

Find the first derivative of the following function at x=0:

$f(x)=e^{x^2\sin(x)}$

Possible Answers:

$2\pi$

Correct answer:

Explanation:

The first derivative of the function is equal to

$f'(x)=(2x\sin(x)+x^2\cos(x))(e^{x^2\sin(x)})$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

Finally, to evaluate the derivative at x=0, simply plug this into the first derivative function:

$f'(0)=(0+0)e^{0}=0$

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Example Question #793 : Other Differential Functions

Find the second derivative of the following function:

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

Possible Answers:

$-16\sin(2x)\cos(2x)$

$-8\sin(2x)\cos(2x)$

$16\sin(2x)\cos(2x)$

$-2\sin^2(2x)+2\cos^2(2x)$

Correct answer:

$-16\sin(2x)\cos(2x)$

Explanation:

The first derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The second derivative - found by taking the derivative of the first derivative function - is equal to

$f''(x)=-16\sin(2x)\cos(2x)$

and was found using the same rules as above.

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Example Question #793 : Other Differential Functions

Find the derivative.

y^2+5y-2

Possible Answers:

2y-5

2y+5

2y^2+5y

Correct answer:

2y+5

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dy}y^2=2y$

$\frac{d}{dy}5y=5$

$\frac{d}{dy}-2=0$

Thus, the derivative is 2y+5 .

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Example Question #982 : Functions

Find the derivative:

f(x)= (9x^2 - 2x + 5)^4

Possible Answers:

$f'(x)= (9x^2 - 2x + 5)^{4} (18x - 2)$

$f'(x)= 4(9x^2 - 2x + 5)^{3} (18x - 2x-5)$

Answer not listed

$f'(x)= 4(9x^2 - 2x + 5)^{3} (18x - 2)$

$f'(x)= 4 (18x - 2)^{3}$

Correct answer:

$f'(x)= 4(9x^2 - 2x + 5)^{3} (18x - 2)$

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)= (9x^2 - 2x + 5)^4

That is done by doing the following: $f'(x)= (4)(9x^2 - 2x + 5)^{4-1} ((2)9x^{2-1} - (1)2x^{1-1} + 0)$

Therefore, the answer is: $f'(x)= 4(9x^2 - 2x + 5)^{3} (18x - 2)$

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Example Question #2012 : Calculus

Find the derivative:

f(x)= 23 + e

Possible Answers:

f'(x)= 0

Answer not listed

f'(x)= 1

f'(x)= e^x

f'(x)= e

Correct answer:

f'(x)= 0

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)= 23 + e

That is done by doing the following: f'(x)= 0 + 0

Therefore, the answer is: f'(x)= 0

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Example Question #2013 : Calculus

Find the derivative:

f(x)= ln(2x) + ln(3x)

Possible Answers:

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

Answer not listed

$f'(x)= \frac{2}{x} + \frac{3}{x}$

$f'(x)= \frac{6}{x}$

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

Correct answer:

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

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)= ln(2x) + ln(3x)

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

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

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Example Question #2014 : Calculus

Find the derivative:

$f(x)= \sin (4x^4 - x)$

Possible Answers:

$f'(x)= (4x^4 - x) (16x^{3} -1)$

$f'(x)= \cos(4x^4 - x) (16x^{3} -1)$

$f'(x)= \cos(4x^4 - x)$

Answer not listed

$f'(x)= \cos(4x^4 - x) \cos(16x^{3} -1)$

Correct answer:

$f'(x)= \cos(4x^4 - x) (16x^{3} -1)$

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: $f(x)= \sin (4x^4 - x)$

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

Therefore, the answer is: $f'(x)= \cos(4x^4 - x) (16x^{3} -1)$

Report an Error

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 #982 : Differential Functions

Find the derivative:

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

Possible Answers:

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

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

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

f'(x)= 1

Answer not listed

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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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #794 : How To Find Differential Functions

Example Question #793 : Other Differential Functions

Example Question #793 : Other Differential Functions

Example Question #982 : Functions

Example Question #2012 : Calculus

Example Question #2013 : Calculus

Example Question #2014 : Calculus

Example Question #2015 : Calculus

Example Question #982 : Differential Functions

Example Question #2016 : Calculus

All Calculus 1 Resources

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