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Calculus 1 : How to find differential functions

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Example Questions

Example Question #1832 : Calculus

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

Find the derivative.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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.

In this case, we must find the derivative of the following:

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

That is done by doing the following:

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

Therefore, the answer is:

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

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Example Question #1833 : Calculus

$f(x)= \sec x - \cos x$

Find the derivative.

Possible Answers:

$f'(x)= \sec x \tan x$

$f'(x)= \tan x +\sin x$

$f'(x)= \sec x \tan x +\cos x$

$f'(x)= \sec x \tan x +\sin x$

$f'(x)= \sec x + \tan x$

Correct answer:

$f'(x)= \sec x \tan x +\sin x$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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.

In this case, we must find the derivative of the following:

$f(x)= \sec x - \cos x$

That is done by doing the following:

$f'(x)= (\sec x) (\tan x) - (-\sin x)$

Therefore, the answer is:

$f'(x)= \sec x \tan x +\sin x$

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Example Question #1834 : Calculus

$f(x)= -\cot x + \tan x + 3x$

Find the derivative.

Possible Answers:

$f'(x)= \csc^2 x + \sec ^2 x + 3$

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

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

$f'(x)= \csc x + \sec x + 3$

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

Correct answer:

$f'(x)= \csc^2 x + \sec ^2 x + 3$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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.

In this case, we must find the derivative of the following:

$f(x)= -\cot x + \tan x + 3x$

That is done by doing the following:

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

Therefore, the answer is:

$f'(x)= \csc^2 x + \sec ^2 x + 3$

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Example Question #811 : Functions

$f(x)= \csc x - 2ln(x) +e^x + 4x^2 -5x$

Find the derivative.

Possible Answers:

$f'(x)= -\csc^2 x - \frac{2}{x} +e^x + 8x -5$

$f'(x)= -\cot x \csc x - \frac{2}{x} +e^x + 8x -5$

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

$f'(x)= \cot x \csc x - \frac{2}{x} +e^x + 8x$

$f'(x)= -\cot x \csc x - \frac{1}{x} +e^x + 8x -5$

Correct answer:

$f'(x)= -\cot x \csc x - \frac{2}{x} +e^x + 8x -5$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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.

In this case, we must find the derivative of the following:

$f(x)= \csc x - 2ln(x) +e^x + 4x^2 -5x$

That is done by doing the following:

$f'(x)= (-\cot x)(\csc x) - (2)(\frac{1}{x}) +(e^x) + (2)4x^{(2-1)} -(1)5x^{(1-1)}$

Therefore, the answer is:

$f'(x)= -\cot x \csc x - \frac{2}{x} +e^x + 8x -5$

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Example Question #811 : Differential Functions

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

Find the derivative.

Possible Answers:

f'(x)= x^2 - x

f'(x)= 3x^2 - 12x^2

f'(x)= 3x^3 - 12x

f'(x)= 3x^2 - 12x + 3

$f'(x)= \frac{1}{3}x^2 - 12x$

Correct answer:

f'(x)= x^2 - x

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

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

That is done by doing the following:

$f'(x)= (3)\frac{1}{3}x^{(3-1)} - (2)\frac{1}{2}x^{(2-1)}+0$

Therefore, the answer is:

f'(x)= x^2 - x

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Example Question #1841 : Calculus

$f(x)= 6x^5+2x^\frac{1}{2}+34x$

Find the derivative.

Possible Answers:

$f'(x)= 5x^4+x^{2}+34$

$f'(x)= 30x^4+x^{-\frac{1}{2}}$

$f'(x)= 30x^5+x^{\frac{1}{2}}+34x$

$f'(x)= 30x^4+x^{-\frac{1}{2}}+34$

$f'(x)= 5x^4+x^{\frac{1}{2}}+34$

Correct answer:

$f'(x)= 30x^4+x^{-\frac{1}{2}}+34$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

$f(x)= 6x^5+2x^\frac{1}{2}+34x$

That is done by doing the following:

$f(x)= (5)6x^{(5-1)}+(\frac{1}{2})2x^{(\frac{1}{2}-1)}+(1)34x^{(1-1)}$

Therefore, the answer is:

$f'(x)= 30x^4+x^{-\frac{1}{2}}+34$

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Example Question #813 : Differential Functions

$f(x)= 6ln(x) - 5e^x + 6x^{-\frac{2}{3}}$

Find the derivative.

Possible Answers:

$f'(x)= \frac{6}{x} - 5e^x - 4 x ^{-\frac{5}{3}}$

$f'(x)= \frac{6}{x} - 5e^x - \frac{2}{3} x ^{-\frac{5}{3}} + 4$

$f'(x)= \frac{6}{x} - e^x - \frac{2}{3} x ^{-\frac{5}{3}}$

$f'(x)= \frac{6}{x} - 5e^x - 4x ^{-\frac{4}{3}}$

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

Correct answer:

$f'(x)= \frac{6}{x} - 5e^x - 4 x ^{-\frac{5}{3}}$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

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

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

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

There are serveral other rules for finding derivatives of different types of functions.

In this case, we must find the derivative of the following:

$f(x)= 6ln(x) - 5e^x + 6x^{-\frac{2}{3}}$

That is done by doing the following:

$f'(x)= (6)(\frac{1}{x}) - (5)(e^x) + (-\frac{2}{3})6x^{(-\frac{2}{3}-1)}$

Therefore, the answer is:

$f'(x)= \frac{6}{x} - 5e^x - 4 x ^{-\frac{5}{3}}$

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Example Question #1842 : Calculus

y = xy^2

Find $\frac{dy}{dx}$ for the follow function:

y = xy^2

Possible Answers:

$y = 2x^{-3}$

$y = -\frac{1}{x}$

$y = -\frac{1}{2x^3}$

$y = -x^{-2}$

Correct answer:

$y = -x^{-2}$

Explanation:

Simplify to make solving for easier:

$y = xy^2 -->\frac{1}{x}=y$

$y = x^{-1}$

To derive this term, take the value of the expnonent and multiply it to . Then subtract 1 from the value of the exponent.

$y = -x^{-2}$

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Example Question #625 : Other Differential Functions

Find $\frac{dy}{dx}$ for the follow function:

y = cos(y) + sin(x)

Possible Answers:

$\frac{dy}{dx} = \frac{cos^2(x)}{1+sin(y)}$

$\frac{dy}{dx} = \frac{cos(x)}{1+sin(y)}$

$\frac{dy}{dx} = \frac{sin(x)}{1+cos(y)}$

$\frac{dy}{dx} = \frac{cos(y)}{1+sin(x)}$

$\frac{dy}{dx} = \frac{-cos(x)}{1+sin(y)}$

Correct answer:

$\frac{dy}{dx} = \frac{cos(x)}{1+sin(y)}$

Explanation:

Take the derivative of both sides of the equation, note that the derivative of y is $\frac{dy}{dx}$ in this case. And remember that the derivative of cos(a) = -sin(a) and that the derivative of sin(a) = cos(a)

$\frac{dy}{dx} = -\frac{dy}{dx}sin(y) + cos(x)$

Now try to separate $\frac{dy}{dx}$

$\frac{dy}{dx}+ \frac{dy}{dx}sin(y)= cos(x) --> \frac{dy}{dx} (1+sin(y)) = cos(x)$

$\frac{dy}{dx} = \frac{cos(x)}{1+sin(y)}$

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Example Question #1841 : Calculus

Find $\frac{dy}{dx}$ for the follow function:

$y = x^{cos(y)}$

Possible Answers:

$\frac{dy}{dx}= \frac{cos(y)}{x( \frac{1}{y}-sin(y)ln(x))}$

$\frac{dy}{dx}= \frac{cos(y)}{x( 1+sin(y)ln(x))}$

$\frac{dy}{dx}= \frac{cos(y)}{x( \frac{1}{y}+sin(y)ln(x))}$

$\frac{dy}{dx}= \frac{xycos(y)}{( \frac{1}{y}+sin(y)ln(x))}$

$\frac{dy}{dx}= \frac{sin(x)}{x( \frac{1}{y}+cos(y))}$

Correct answer:

$\frac{dy}{dx}= \frac{cos(y)}{x( \frac{1}{y}+sin(y)ln(x))}$

Explanation:

To make solving easier, take the natural log of both sides of the equations:

$ln(y) = ln(x^{cos(y)})$

Use the natural log properties of ln(a^b) = bln(a) .

ln(y) = cos(y)ln(x)

Now take the derivative of both sides of the equation, note that the derivative of is $\frac{dy}{dx}$ in this case. Must also use the power rule for cos(y)ln(x) . The general equation is $\frac{d(ab)}{dx}= \frac{da}{dx}b+a\frac{db}{dx}$

$\frac{dy}{dx} \frac{1}{y}= -\frac{dy}{dx}sin(y)ln(x)+\frac{cos(y)}{x}$

Now, isolate for $\frac{dy}{dx}$ :

$\frac{dy}{dx} \frac{1}{y}+\frac{dy}{dx}sin(y)ln(x)= \frac{cos(y)}{x}$

$\frac{dy}{dx}( \frac{1}{y}+sin(y)ln(x))= \frac{cos(y)}{x}$

$\frac{dy}{dx}= \frac{cos(y)}{x( \frac{1}{y}+sin(y)ln(x))}$

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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 #1832 : Calculus

Example Question #1833 : Calculus

Example Question #1834 : Calculus

Example Question #811 : Functions

Example Question #811 : Differential Functions

Example Question #1841 : Calculus

Example Question #813 : Differential Functions

Example Question #1842 : Calculus

Example Question #625 : Other Differential Functions

Example Question #1841 : Calculus

All Calculus 1 Resources

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