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Calculus 1 : How to find differential functions

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Example Questions

Example Question #671 : Other Differential Functions

Find the derivative of the following: f(x) = ln(2x^3 + e^x)

Possible Answers:

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

$f'(x) = \frac{6x^2+e^x}{e^x}$

$f'(x) = - \frac{6x^2+e^x}{2x^3 + e^x}$

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

$f'(x) = ln \frac{6x^2+e^x}{2x^3 + e^x}$

Correct answer:

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

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

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

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

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Example Question #672 : Other Differential Functions

Find the derivative of the following: $f(x) = e^{4x}$

Possible Answers:

$f'(x) = e^{4x}$

$f'(x) = 4e^{3}$

$f'(x) = 4e^{4x}$

$f'(x) = 4e^{x}$

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

Correct answer:

$f'(x) = 4e^{4x}$

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: $f(x) = e^{4x}$

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

Therefore, the answer is: $f'(x) = 4e^{4x}$

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Example Question #673 : Other Differential Functions

Find the derivative of the following: $f(x) = \sin (ln(2x))$

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #861 : Functions

Determine the slope of the line that is tangent to the function f(x)=ln(ln(cos(x))) at the point x=1.5

Possible Answers:

34.78

-0.38

-5.32

-34.78

5.32

Correct answer:

5.32

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(ln(cos(x))) at the point x=1.5

The slope of the tangent is

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

$f'(1.5)=\frac{-tan(1.5)}{ln(cos(1.5))}=5.32$

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Example Question #675 : Other Differential Functions

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

Possible Answers:

$\frac{2}{ln(2)}$

$\sqrt{2}$

$-\frac{\sqrt{2}}{2}$

$-\frac{2}{ln(2)}$

$\frac{\sqrt{2}}{2}$

Correct answer:

$-\frac{2}{ln(2)}$

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

The slope of the tangent is

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

$f'(\frac{\pi}{4})=\frac{cot(\frac{\pi}{4})}{ln(sin(\frac{\pi}{4}))}=-\frac{2}{ln(2)}$

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Example Question #676 : Other Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=cos(x^2+4x+\frac{\pi}{2})$ at the point x=0

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.

We'll need to make use of the following derivative rule(s):

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 $f(x)=cos(x^2+4x+\frac{\pi}{2})$ at the point x=0

The slope of the tangent is

$f'(x)=-(2x+4)sin(x^2+4x+\frac{\pi}{2})$

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

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Example Question #862 : Functions

Determine the slope of the line that is tangent to the function $f(x)=sin(\pi x+\pi)$ at the point x=3

Possible Answers:

$-\pi$

$3\pi$

$-4\pi$

$4\pi$

$\pi$

Correct answer:

$\pi$

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 $f(x)=sin(\pi x+\pi)$ at the point x=3

The slope of the tangent is

$f'(x)=\pi cos(\pi x+\pi)$

$f'(3)=\pi cos(3\pi +\pi)=\pi$

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Example Question #678 : Other Differential Functions

Determine the slope of the line that is tangent to the function f(x)=x^6+3x^4+8x^2+10 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)=x^6+3x^4+8x^2+10 at the point x=1

The slope of the tangent is

f'(x)=6x^5+12x^3+16x

f'(1)=6(1)^5+12(1)^3+16(1)=34

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Example Question #679 : Other Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=cos(\pi(x^2+6x+4))$ at the point x=2

Possible Answers:

$-\frac{7\sqrt{2}}{2}\pi$

$\frac{7\sqrt{2}}{2}\pi$

$-\frac{7}{2}\pi$

$7\sqrt{2}\pi$

$\frac{7}{2}\pi$

Correct answer:

$\frac{7\sqrt{2}}{2}\pi$

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

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

Taking the derivative of the function $f(x)=cos(\pi(x^2+6x+4))$ at the point $x=\frac{1}{2}$

The slope of the tangent is

$f'(x)=-\pi(2x+6)sin(\pi(x^2+6x+4))$

$f'(\frac{1}{2})=-\pi(2(\frac{1}{2})+6)sin(\pi(\frac{1}{2}^2+6(\frac{1}{2})+4))=\frac{7\sqrt{2}}{2}\pi$

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Example Question #680 : Other Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=e^{sin(\pi x^2)}$ at the point x=3

Possible Answers:

$-3\pi$

$-6\pi$

$3\pi$

$6\pi$

Correct answer:

$-6\pi$

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 an exponential:

d[e^u]=e^udu

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 $f(x)=e^{sin(\pi x^2)}$ at the point x=3

The slope of the tangent is

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

$f'(3)=2\pi (3)cos(\pi (3)^2)e^{sin(\pi (3)^2)}=-6\pi$

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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 #671 : Other Differential Functions

Example Question #672 : Other Differential Functions

Example Question #673 : Other Differential Functions

Example Question #861 : Functions

Example Question #675 : Other Differential Functions

Example Question #676 : Other Differential Functions

Example Question #862 : Functions

Example Question #678 : Other Differential Functions

Example Question #679 : Other Differential Functions

Example Question #680 : Other Differential Functions

All Calculus 1 Resources

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