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Calculus AB : Differentiating Functions

Study concepts, example questions & explanations for Calculus AB

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Example Questions

Example Question #1 : Differentiate Inverse Functions

Which expression correctly identifies the inverse of f (x) = 2x-4 ?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The inverse of a function can be found by substituting yvariables for the variables found in the function, then setting the function equal to . By next isolating , the inverse function is written. Then, the notation $f^{-1}$ is used to describe the newly written function as being the inverse of the original function. The answer choice “ $f^{-1}(x) =\frac{1}{2}x+2$ ” is correct.

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Example Question #211 : Calculus Ab

Which of the following correctly identifies the derivative of an inverse function?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

This question asks you to recognize the correct notation of a differentiating inverse functions problem. First, it is key to recognize that the equation needs to have the same variable throughout, thus eliminating the answer choices $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(x)) }$ and $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(y))}$ . Next, there should be no constants in the correct equation; thus, $(f^{-1})'(x) = \frac{2}{f'(f^{-1}(x))}$ is incorrect. The correct choice is $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

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Example Question #1 : Differentiate Inverse Functions

Find $(f^{-1})'(1)$ given f (x) = x^5+2x+1 .

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{2}{3}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{2}$

Explanation:

Let f (x) = x^5+2x+1

It is important to recognize the relationship between a function and its inverse to solve.

If f (0) = (0)^5+2(0)+1=1 , solving for the inverse function will produce $f^{-1}(1)=0$ .

To find the derivative of an inverse function, use: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

$(f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))}=\frac{1}{f' (0)}=\frac{1}{5(0)^4+ 2}=\frac{1}{0 + 2}=\frac{1}{2}$

Therefore, $(f^{-1})'(1)=\frac{1}{2}$

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Example Question #2 : Differentiate Inverse Functions

Let f (x) =x^6 .Find $(f^{-1})'(x)$ .

Possible Answers:

$\frac{6}{x}$

$\frac{1}{6x}$

$\frac{1}{6x^5_6}$

$\frac{1}{4x^5_6}$

Correct answer:

$\frac{1}{6x^5_6}$

Explanation:

To find the derivative of the inverse of f (x) , it is useful to first solve for $f^{-1}$ .

This will help because $f^{-1}$ is needed in the derivative equation, $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

$f (x) =x^6 \rightarrow f^{-1} (x) =\sqrt[6]{x}$

Next, the equation for the derivative of an inverse function can be evaluated.

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[6]{x})}$

After substituting in $\sqrt[6]{x}$ for $f^{-1} (x)$ , the derivative of f (x), or f'(x) =6x^5 , is applied:

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[6]{x})}=\frac{1}{6(\sqrt[6]{x})^5}=\frac{1}{6x^5_6}$

Therefore, the correct answer is $(f^{-1})'(x)=\frac{1}{6x^5_6}$

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Example Question #1 : Differentiate Inverse Functions

Let f (x) =e^x . Find $(f^{-1})'(x)$ .

Possible Answers:

ln(x)

$\frac{1}{x}$

$xe^{x-1}$

e^x

Correct answer:

$\frac{1}{x}$

Explanation:

To find the derivative of the inverse of f (x) , it is useful to first solve for $f^{-1}$ .

This will help because $f^{-1}$ is needed in the derivative equation, $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

$f (x) =e^x \rightarrow f^{-1} (x) =ln(x)$

Next, the equation for the derivative of an inverse function can be evaluated.

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

After substituting in $ln(x) for f^{-1} (x)$ , the derivative of f (x) , or f'(x) =e^x , is applied:

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

Therefore, the correct answer is $(f^{-1})'(x)=\frac{1}{x}$

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Example Question #2 : Differentiate Inverse Functions

Let $f (x) =\sqrt[3]{x}$ . Find $(f^{-1})'(x)$ .

Possible Answers:

2x^3

$\frac{3}{x^2}$

$\frac{1}{x^2}$

3x^2

Correct answer:

3x^2

Explanation:

To find the derivative of the inverse of f (x) , it is useful to first solve for $f^{-1}$ .

This will help because $f^{-1}$ is needed in the derivative equation, $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

$f (x)=\sqrt[3]{x} \rightarrow f^{-1} (x)=x^3$

Next, the equation for the derivative of an inverse function can be evaluated.

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

After substituting in x^3 for $f^{-1} (x)$ , the derivative of f (x) , or $f'(x)=(\frac{1}{3})(x^{-\frac{2}{3}})$ , is applied:

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

Therefore, the correct answer is $(f^{-1})'(x)=3x^2$ .

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Example Question #212 : Calculus Ab

Let f (x) =2-2x . Find $(f^{-1})'(x)$ .

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{3}$

Correct answer:

$-\frac{1}{2}$

Explanation:

To find the derivative of the inverse of f (x) , it is useful to first solve for $f^{-1}$ .

This will help because $f^{-1}$ is needed in the derivative equation, $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

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

Next, the equation for the derivative of an inverse function can be evaluated.

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

After substituting in $1-\frac{1}{2}x$ for $f^{-1} (x)$ , the derivative of f (x) , or f'(x)= -2 , is applied:

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

Therefore, the correct answer is $(f^{-1})'(x)=-\frac{1}{2}$ .

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Example Question #1 : Differentiate Inverse Functions

Let f (x) =6x^2 . Find $(f^{-1})'(x)$ .

Possible Answers:

$\frac{1}{12(\sqrt{}^x_6)}$

$\frac{1}{2x}$

$\frac{1}{12x^2}$

Correct answer:

$\frac{1}{12(\sqrt{}^x_6)}$

Explanation:

To find the derivative of the inverse of f (x) , it is useful to first solve for $f^{-1}$ .

This will help because $f^{-1}$ is needed in the derivative equation, $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

$f (x) =6x^2 \rightarrow f^{-1} (x) =\sqrt[2]{^x_6}$

Next, the equation for the derivative of an inverse function can be evaluated.

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[2]{^x_6})}$

After substituting in $\sqrt[2]{^x_6}$ for $f^{-1} (x)$ , the derivative of f (x) , or f'(x) =12x , is applied:

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[2]{^x_6})}=\frac{1}{12(\sqrt[2]{^x_6})}$

Therefore, the correct answer is $(f^{-1})'(x)=\frac{1}{12(\sqrt{^x_6})}$

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Example Question #2 : Differentiate Inverse Functions

Suppose the points in the table below represent the continuous function f(x) . The differentiable function g(x) is the inverse of the function f(x) . Find g'(2) .

Q9 table

Possible Answers:

$\frac{1}{6}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{6}$

Explanation:

Below is the equation for the derivative of g(x) :

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

$g'(2) = \frac{1}{f'(f^{-1}(2))}$

So, the value of $f^{-1}(2)$ must first be found.

Using the data from the table, $f^{-1}(2)=2$ since f(2)=2 .

Next, from the table the following can be obtained:

f'(2)=6

Now, the appropriate substitutions can be made to solve for g'(2) .

$g'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(2)}=\frac{1}{6}$

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Example Question #1 : Differentiate Inverse Functions

Find $(f^{-1})'(3)$ given f (x) =ln(3x) .

Possible Answers:

e^9

$\frac{1}{3}e^9$

e^3

$\frac{1}{3}e^3$

Correct answer:

$\frac{1}{3}e^3$

Explanation:

To find the derivative of the inverse of f (x) , it is useful to first solve for $f^{-1}$ .

This will help because $f^{-1}$ is needed in the derivative equation, $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ .

$f (x) =ln(3x) \rightarrow f^{-1} (x) =\frac{1}{3}e^x$

Next, the equation for the derivative of an inverse function can be evaluated.

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

After substituting in $\frac{1}{3}e^x$ for $f^{-1} (x)$ , the derivative of f (x) , or $f'(x) =\frac{1}{x}$ , is found by taking the derivative of ln(3x) and applying chain rule.

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

After finding the general term $(f^{-1})'(x)=\frac{1}{3}e^x$ , evaluate at x=3 .

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

Therefore, the correct answer is $(f^{-1})'(3)=\frac{1}{3}e^3$ .

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Calculus AB : Differentiating Functions

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Differentiate Inverse Functions

Example Question #211 : Calculus Ab

Example Question #1 : Differentiate Inverse Functions

Example Question #2 : Differentiate Inverse Functions

Example Question #1 : Differentiate Inverse Functions

Example Question #2 : Differentiate Inverse Functions

Example Question #212 : Calculus Ab

Example Question #1 : Differentiate Inverse Functions

Example Question #2 : Differentiate Inverse Functions

Example Question #1 : Differentiate Inverse Functions

All Calculus AB Resources

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