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

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 \(\displaystyle (f^{-1})'(y) = \frac{1}{f'(f^{-1}(x)) }\)and \(\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(y))}\). Next, there should be no constants in the correct equation; thus, \(\displaystyle (f^{-1})'(x) = \frac{2}{f'(f^{-1}(x))}\) is incorrect. The correct choice is \(\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Next, from the table the following can be obtained: 

\(\displaystyle f'(2)=6\)

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

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

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

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

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

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

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

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

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

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

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

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