Question

Use the formula in Exercise 83.

85. If \(f\left( x \right) = x + {e^x}\), find \({\left( {{f^{ - 1}}} \right)^\prime }\left( 1 \right)\).

Short answer

The value is \(\frac{1}{2}\).

Step-by-step solution

The derivative of the inverse function

If \(f\left( x \right)\) be a differentiable and invertible function, then the derivative of \({f^{ - 1}}\) can be obtained using the formula: \({\left( {{f^{ - 1}}} \right)^\prime }\left( x \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}}\).

The derivative of the inverse function

It is given that \(f\left( x \right) = x + {e^x}\) that implies \({f^{ - 1}}\left( x \right) = 1 + {e^x}\). So, \(f\left( 0 \right) = 0 + 1 = 1\) that implies \({f^{ - 1}}\left( 1 \right) = 0\) and \(f'\left( 0 \right) = 1 + 1 = 2\).

Substitute \(x = 1\) in the above formula:

\(\begin{array}{l}{\left( {{f^{ - 1}}} \right)^\prime }\left( 1 \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( 1 \right)} \right)}}\\{\left( {{f^{ - 1}}} \right)^\prime }\left( 1 \right) = \frac{1}{{f'\left( 0 \right)}}\\{\left( {{f^{ - 1}}} \right)^\prime }\left( 1 \right) = \frac{1}{2}\end{array}\)

Hence, \({\left( {{f^{ - 1}}} \right)^\prime }\left( 1 \right) = \frac{1}{2}\).