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Calculus 1 : Differential Functions

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Example Questions

Example Question #612 : How To Find Differential Functions

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

y^2 + y +x =0

Hint: Use implicit differentiation.

Possible Answers:

$\frac{dy}{dx}={2y+x+1}$

$\frac{dy}{dx}= x +2$

$\frac{dy}{dx}=\frac{-1}{2x+1}$

$\frac{dy}{dx}=\frac{-1}{2y+1}$

$\frac{dy}{dx}= y^2+1$

Correct answer:

$\frac{dy}{dx}=\frac{-1}{2y+1}$

Explanation:

Take the derivative of each term. Remember that the derivative of y should be dy/dx.

$\frac{d(y^2 + y +x =0)}{dx} = 2y\frac{dy}{dx}+\frac{dy}{dx}+1=0$

Remember to follow through with the chain rule for y^2 . Have to multiply by the derivative of the inside of the squared, which is just $\frac{dy}{dx}$ .

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

$\frac{dy}{dx}(2y+1)=-1$

$\frac{dy}{dx}=\frac{-1}{2y+1}$

Report an Error

Example Question #801 : Differential Functions

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

y = x^x

Hint: Use implicit differentiation.

Possible Answers:

$\frac{dy}{dx}= y(lnx)^2$

$\frac{dy}{dx}= (xlnx+x)$

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

$\frac{dy}{dx}= x+lnx$

$\frac{dy}{dx}= y(lnx+1)$

Correct answer:

$\frac{dy}{dx}= y(lnx+1)$

Explanation:

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

ln(y) = ln(x^x)

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

ln(y) = xln(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 the xln(x) term. The general equation is $\frac{d(ab)}{dx}$ = $\frac{da}{dx}*b+a*\frac{db}{dx}$

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

Now simplify:

$\frac{dy}{dx}= y(lnx+1)$

Report an Error

Example Question #1831 : Calculus

$f(x)= 8x^{\frac{1}{8}} - 12x^{\frac{1}{2}} + 4x - 6$

Find the derivative.

Possible Answers:

$f'(x)= 8x^{-\frac{7}{8}} - 12x^{-\frac{1}{2}}+4$

$f'(x)= x^{-\frac{7}{8}} - 6x^{-\frac{1}{2}}+4$

$f'(x)= x^{\frac{7}{8}} - 6x^{\frac{1}{2}}+4$

$f'(x)= x^{-\frac{7}{8}} - 6x^{-\frac{1}{2}}$

$f'(x)= x^{-\frac{7}{8}} - 6x^{-\frac{1}{2}}+4 +x$

Correct answer:

$f'(x)= x^{-\frac{7}{8}} - 6x^{-\frac{1}{2}}+4$

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:

$f(x)= 8x^{\frac{1}{8}} - 12x^{\frac{1}{2}} + 4x - 6$

That is done by doing the following:

$f'(x)= (\frac{1}{8})8x^{(\frac{1}{8}-1)} - (\frac{1}{2})12x^{(\frac{1}{2}-1)} + (1)4x^{(1-1)} - 0$

Therefore, the answer is:

$f'(x)= x^{-\frac{7}{8}} - 6x^{-\frac{1}{2}}+4$

Report an Error

Example Question #611 : How To Find Differential Functions

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

Find the derivative.

Possible Answers:

$f'(x)= 2x^{-\frac{1}{3}} - 4x^4$

$f'(x)= 2x^{-\frac{1}{3}} - 4x^3 + 4$

$f'(x)= 2x^{-\frac{1}{3}} - 4x^3$

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

$f'(x)= 2x^{\frac{1}{3}} - 4x^3$

Correct answer:

$f'(x)= 2x^{-\frac{1}{3}} - 4x^3$

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 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:

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

That is done by doing the following:

$f'(x)= (\frac{2}{3})3x^{(\frac{2}{3}-1)} - (4)x^{(4-1)} + 0$

Therefore, the answer is:

$f'(x)= 2x^{-\frac{1}{3}} - 4x^3$

Report an Error

Example Question #612 : Other Differential Functions

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

Find the derivative.

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

That is done by doing the following:

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

Therefore, the answer is:

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

Report an Error

Example Question #613 : Other Differential Functions

f(x)= 2ln(x) - 8x^3 +4x^6 -23

Find the derivative.

Possible Answers:

$f'(x)= \frac{2}{x^2} - 24x^2 + 24x^5$

$f'(x)= \frac{2}{x} - 24x^2 + 24x^5 +6$

$f'(x)= \frac{2}{x} - 24x^2 + 24x^5$

f'(x)= x - 24x^2 + 24x^5

$f'(x)= \frac{2}{x} - 3x^2 + 24x^5$

Correct answer:

$f'(x)= \frac{2}{x} - 24x^2 + 24x^5$

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

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)= 2ln(x) - 8x^3 +4x^6 -23

That is done by doing the following:

$f'(x)= 2(\frac{1}{x}) - (3)8x^{(3-1)} +(6)4x^{(6-1)} -0$

Therefore, the answer is:

$f'(x)= \frac{2}{x} - 24x^2 + 24x^5$

Report an Error

Example Question #614 : Other Differential Functions

$f(x)= 3x^2 - 4ln(x) + 100x^{\frac{1}{100}}$

Find the derivative.

Possible Answers:

$f'(x)= 6x - \frac{4}{x} + x^{-\frac{99}{100}}$

$f'(x)= 6x - \frac{1}{x} + x^{-\frac{99}{100}}$

$f'(x)= 6x^2 - \frac{4}{x} + x^{-\frac{99}{100}}$

$f'(x)= x - \frac{1}{x} + 100x^{-\frac{99}{100}}$

$f'(x)= 6x - \frac{4}{x} + 100x$

Correct answer:

$f'(x)= 6x - \frac{4}{x} + x^{-\frac{99}{100}}$

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:

$f(x)= 3x^2 - 4ln(x) + 100x^{\frac{1}{100}}$

That is done by doing the following:

$f'(x)= (2)3x^{(2-1)} - (4)(\frac{1}{x}) + (\frac{1}{100})100x^{(\frac{1}{100}-1)}$

Therefore, the answer is:

$f'(x)= 6x - \frac{4}{x} + x^{-\frac{99}{100}}$

Report an Error

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:

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

Report an Error

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:

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

Report an Error

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:

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

Report an Error

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #612 : How To Find Differential Functions

Example Question #801 : Differential Functions

Example Question #1831 : Calculus

Example Question #611 : How To Find Differential Functions

Example Question #612 : Other Differential Functions

Example Question #613 : Other Differential Functions

Example Question #614 : Other Differential Functions

Example Question #1832 : Calculus

Example Question #1833 : Calculus

Example Question #1834 : Calculus

All Calculus 1 Resources

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