Derivative of the inverse of a function

Summary

• \(f:A→B\) is an arbitrary smooth function.
• \(x_0\) is an arbitrary point in \(A\) such that \(f'\left( x_0 \right)≠0\) and \(y_0=f\left( x_0 \right)\) .
• By the inverse function theorem, there exists an open interval \(\left( a,b \right)\) containing \(x_0\) such that \(f\) restricted to the interval such that \(f\) has an inverse.
• Denote this inverse by \(x=g\left( y \right)\) .
• Derivative of the inverse:

\[g'\left( y_0 \right)= \frac{1}{f'\left( x_0 \right)}\]

• Example: \(f\left( x \right)=x^2\) , \(x_0=1\) and \(y_0=f\left( x_0 \right)=1^2=1\) .
• Since \(f'\left( x \right)=2x\) , \(f'\left( x_0 \right)=2≠0\) , the inverse function theorem applies and

\[g'\left( y_0 \right)= \frac{1}{f'\left( x_0 \right)}= \frac{1}{2}\]

• Check: the inverse function of \(f\) around \(x=1\) is \(g\left( y \right)=\sqrt{y}\) .
• \(g'\left( y \right)= \frac{1}{2\sqrt{y}}\) and \(g'\left( y_0 \right)= \frac{1}{2\sqrt{1}}= \frac{1}{2}\) .