There are a few methods to solve this problem.

1. Method 1: Find the inverse\[\] First, find the inverse of the function. \[y=x^{2}-5\] \[\mbox{Swap x and y}\] \[x=y^{2}-5\] \[\mbox{Solve for y}\] \[x+5=y^{2}\] \[y=\pm\sqrt{x+5}\] \[\therefore\] \[f^{-1}\left(x\right)=\pm\sqrt{x+5}\] Next, take the derivative of the inverse function like you would another derivative. \[\left[f^{-1}\left(x\right)\right]^{'}=\pm\ \frac{d}{dx}\left[\left(x+5\right)^{\frac{1}{2}}\right]=\pm\ \frac{1}{2}\left(x+5\right)^{-\frac{1}{2}}=\] \[\pm\ \frac{1}{2\sqrt{x+5}}\]
2. Method 2: Using the Formula \[\ First, take the derivative. \[f\left(x\right)=x^{2}-5\\] \[f'\left(x\right)=2x\] Next, swap \(x\) and \(y\) in the derivative function. \[f'\left(x\right)=y'=2x\] \[x'=2y\] \[f'\left(y\right)=2y\] Next, subsitute \(f'\left(y\right)\) into the formula. \[\left[f^{-1}\left(x\right)\right]^{'}=\frac{1}{2y}\] Now, the above expression is the derivative in terms of \(y\); however, to get it in terms of \(x\) we can't just substitute the original function in for \(y\). That's because we need to remember that in the formula \(y=f^{-1}\left(x\right)\); therefore, we need to substitute the inverse of the function in for \(y\). \[\mbox{Find the inverse using the same approach as in method 1}\] \[f^{-1}\left(x\right)=\pm\sqrt{x+5}\] Lastly, substitute the above expression into the derivative for \(y\) \[\left[f^{-1}\left(x\right)\right]^{'}=\frac{1}{2y}=\frac{1}{\pm2\sqrt{x+5}}\]

In the previous example we looked at two ways to solve these types of problems. In this example we will use a third method: implicit differentiation.

Swap \(x\) and \(y\) \[y=x^{2}+4x+4\] \[x=y^{2}+4y+4\] Find the derivative using implicit differentiation with respect to \(x\) \[\frac{d}{dx}\left[x\right]=\frac{d}{dx}\left[y^{2}+4y+4\right]\] \[1=2y\cdot\frac{dy}{dx}+4\cdot\frac{dy}{dx}\] \[1=\frac{dy}{dx}\left(2y+4\right)\] \[\frac{dy}{dx}=\frac{1}{2y+4}\] The above expresion is the derivative of the inverse; however, it is in terms of \(y\). To get it in terms of \(x\), we need to substitute the inverse function in for \(y\); therefore, let's find the inverse. \[y=x^{2}+4x+4\] \[x=y^{2}+4y+4=\left(y+2\right)^{2}\] \[\left(y+2\right)^{2}=x\] \[y+2=\pm\sqrt{x}\] \[y=\pm\sqrt{x}-2\] We can substitute the following inverse function in for \(y\) \[\left[f^{-1}\left(x\right)\right]^{'}=\frac{1}{2\left(\pm\sqrt{x}-2\right)+4}=\frac{1}{\pm\ 2\sqrt{x}-4+4}=\frac{1}{\pm\ 2\sqrt{x}}\]

We will use method 1 to solve this problem.

1. Find the inverse \[y=\ln x\] \[x=\ln y\] \[y=e^{x}\] \[f^{-1}\left(x\right)=e^{x}\]
2. Take the derivative of the inverse \[\left[f^{-1}\left(x\right)\right]^{'}=e^{x}\]
3. Evaluate the derivative at \(x=1\) \[\left[f^{-1}\left(1\right)\right]^{'}=e\] \[\therefore\] The rate of change at \(x=1\) is \(e\).

1. Find the inverse of the function \[y=4x^{3}+2\] \[x=4y^{3}+2\] \[x-2=4y^{3}\] \[y^{3}=\frac{x-2}{4}\]\[y=\sqrt[3]{\frac{x-2}{4}}\]
2. Take the derivative of the inverse \[y'=\frac{d}{dx}\left[\sqrt[3]{\frac{x-2}{4}}\right]=\frac{d}{dx}\left[\left(\frac{x-2}{4}\right)^{-\frac{1}{3}}\right]=\] \[-\frac{1}{3}\left(\frac{x-2}{4}\right)^{-\frac{1}{3}-1}=-\frac{1}{3}\left(\frac{x-2}{4}\right)^{-\frac{2}{3}}\]

1. Swap the variables \[y=\frac{2}{x^{2}+1}\] \[x=\frac{2}{y^{2}+1}\]
2. Use implicit differentiation to find the derivative \[\frac{d}{dx}\left[x\right]=\frac{d}{dx}\left[\frac{2}{y^{2}+1}\right]\] \[1=\frac{d}{dx}\left[2\left(y^{2}+1\right)^{-1}\right]\] \[1=-2\left(y^{2}-1\right)^{-2}\left(2y\right)\cdot\frac{dy}{dx}\] \[1=-\frac{4y}{\left(y^{2}-1\right)^{2}}\cdot\frac{dy}{dx}\] \[\frac{dy}{dx}=-\frac{\left(y^{2}-1\right)^{2}}{4y}\]
3. Find the inverse of the original function \[y=\frac{2}{x^{2}+1}\] \[x=\frac{2}{y^{2}+1}\] \[\left(y^{2}+1\right)x=2\] \[y^{2}+1=\frac{2}{x}\] \[y^{2}=\frac{2}{x}+1\] \[y=\pm\sqrt{\frac{2}{x}+1}\]
4. Substitute the above expression in for \(y\) \[\frac{dy}{dx}=-\frac{\left(y^{2}-1\right)^{2}}{4y}=-\frac{\left(\left(\pm\sqrt{\frac{2}{x}+1}\right)^{2}-1\right)^{2}}{4\left(\pm\sqrt{\frac{2}{x}+1}\right)}=\] \[-\frac{\left(\frac{2}{x}+1-1\right)^{2}}{\pm4\sqrt{\frac{2}{x}+1}}=-\frac{\frac{4}{x^{2}}}{\pm4\sqrt{\frac{2}{x}+1}}=\] \[-(\pm)\frac{1}{x^{2}\sqrt{\frac{2}{x}+1}}=\mp\frac{\sqrt{\frac{2}{x}+1}}{x^{2}\left(\sqrt{\frac{2}{x}+1}\right)\left(\sqrt{\frac{2}{x}+1}\right)}=\] \[\mp \frac{\sqrt{\frac{2}{x}+1}}{x^{2}\left(\frac{2}{x}+1\right)}=\mp \frac{\sqrt{\frac{2}{x}+1}}{2x+x^{2}}=\mp\frac{\sqrt{\frac{2}{x}+1}}{x^{2}+2x}\]

\[\therefore\]

\[\left[f^{-1}\left(x\right)\right]^{'}=\mp \frac{\sqrt{\frac{2}{x}+1}}{x^{2}+2x}\]

We will use method 2 to solve this problem.

1. Take the derivative \[f\left(x\right)=x^{2}+6x+9\] \[f'\left(x\right)=2x+6\]
2. Swap \(x\) and \(y\) \[f'\left(y\right)=2y+6\]
3. Substitute the above expresion into the formula \[\left[f^{-1}\left(x\right)\right]^{'}=\frac{1}{2y+6}\]
4. Find the inverse of the function \[f\left(x\right)=x^{2}+6x+9\] \[y=x^{2}+6x+9\] \[x=y^{2}+6y+9\] \[x=\left(y+3\right)^{2}\] \[\pm\sqrt{x}=y+3\] \[y=f^{-1}\left(x\right)=\pm\sqrt{x}-3\]
5. Substitute the inverse in for \(y\) \[\left[f^{-1}\left(x\right)\right]^{'}=\frac{1}{2\left(\pm\sqrt{x}-3\right)+6}=\frac{1}{\pm\ 2\sqrt{x}}\]
6. Find the value of the inverse function at \(x=4\) \[f^{-1}\left(4\right)=\pm\sqrt{4}-3=\pm\ 2-3\]
7. Find the value of the derivative at \(x=4\) \[\left[f^{-1}\left(4\right)\right]^{'}=\frac{1}{\pm\ 2\sqrt{4}}=\pm\ \frac{1}{4}\]
8. Write the equation of the two tangent lines \[\] The \(y\) value of \(+2-3\) will correlate with the slope of \(+\frac{1}{4}\), and the \(y\) value of \(-2-3\) will correlate with the slope of \(-\frac{1}{4}\). \[\therefore\] The first tangent line: \[y-\left(+2-3\right)=\frac{1}{4}\left(x-4\right)\] \[y+1=\frac{1}{4}x-1\] \[y=\frac{1}{4}x-2\] The second tangent line: \[y-\left(-2-3\right)=-\frac{1}{4}\left(x-4\right)\] \[y+5=-\frac{1}{4}x+1\] \[y=-\frac{1}{4}x-4\]

1. Find the inverse function \[\] \[y=2\sqrt{x}+3\] \[x=2\sqrt{y}+3\] \[x-3=2\sqrt{y}\] \[\sqrt{y}=\frac{x-3}{2}\] \[y=\left(\frac{x-3}{2}\right)^{2}\] The above ewaution is equal to \(f^{-1}\left(x\right)\); therefore, it is equal to \(g'\left(x\right)\) \[f^{-1}\left(x\right)=g\left(x\right)=y=\left(\frac{x-3}{2}\right)^{2}\]
2. Take the derivative of \(g(x)\) \[g'\left(x\right)=2\left(\frac{x-3}{2}\right)^{2-1}\cdot\frac{d}{dx}\left[\frac{x-3}{2}\right]=\] \[2\left(\frac{x-3}{2}\right)^{1}\cdot\frac{1}{2}=\frac{x-3}{2}\]
3. Find \(g'(5)\) \[g'\left(5\right)=\frac{5-3}{2}=\frac{2}{2}=1\]
4. Write the equation of the tangent line \[\] We don't have to necessarily find the value of \(g\) at \(x=5\) since we are given \(f(1)=5\). Since \(g\) is the inverse of \(f\), \(g(5)\) is the inverse of \(f(1)\); therefore, \(g(5)=1\). Now we can write the equation. \[y-1=1\left(x-5\right)\] \[y=x-5+1\] \[y=x-4\]