Editing Chain rule (section)

=== Derivatives of inverse functions ===
{{Main|Inverse functions and differentiation}}
Suppose that {{math|1=''y'' = ''g''(''x'')}} has an [[inverse function]]. Call its inverse function {{mvar|f}} so that we have {{math|1=''x'' = ''f''(''y'')}}. There is a formula for the derivative of {{mvar|f}} in terms of the derivative of {{mvar|g}}. To see this, note that {{mvar|f}} and {{mvar|g}} satisfy the formula
<math display="block">f(g(x)) = x.</math>

And because the functions <math>f(g(x))</math> and {{mvar|x}} are equal, their derivatives must be equal. The derivative of {{mvar|x}} is the constant function with value 1, and the derivative of <math>f(g(x))</math> is determined by the chain rule. Therefore, we have that:
<math display="block">f'(g(x)) g'(x) = 1.</math>

To express {{mvar|f'}} as a function of an independent variable {{mvar|y}}, we substitute <math>f(y)</math> for {{mvar|x}} wherever it appears. Then we can solve for {{mvar|f'}}.
<math display="block">\begin{align}
f'(g(f(y))) g'(f(y)) &= 1 \\
f'(y) g'(f(y)) &= 1 \\
f'(y) = \frac{1}{g'(f(y))}.
\end{align}</math>

For example, consider the function {{math|1=''g''(''x'') = ''e''<sup>''x''</sup>}}. It has an inverse {{math|1=''f''(''y'') = ln ''y''}}. Because {{math|1=''g''′(''x'') = ''e''<sup>''x''</sup>}}, the above formula says that
<math display="block">\frac{d}{dy}\ln y = \frac{1}{e^{\ln y}} = \frac{1}{y}.</math>

This formula is true whenever {{mvar|g}} is differentiable and its inverse {{mvar|f}} is also differentiable. This formula can fail when one of these conditions is not true. For example, consider {{math|1=''g''(''x'') = ''x''<sup>3</sup>}}. Its inverse is {{math|1=''f''(''y'') = ''y''<sup>1/3</sup>}}, which is not differentiable at zero. If we attempt to use the above formula to compute the derivative of {{mvar|f}} at zero, then we must evaluate {{math|1=1/''g''′(''f''(0))}}. Since {{math|1=''f''(0) = 0}} and {{math|1=''g''′(0) = 0}}, we must evaluate 1/0, which is undefined. Therefore, the formula fails in this case. This is not surprising because {{mvar|f}} is not differentiable at zero.