Question

Use the formula in Exercise 83.

86. If \(f\left( x \right) = {x^3} + 3\sin x + 2\cos x\), find \({\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right)\).

Short answer

The function is \(\frac{1}{3}\).

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 given inverse

It is given that \(f\left( x \right) = {x^3} + 3\sin x + 2\cos x\) that implies \(f'\left( x \right) = 3{x^2} + 3\cos x - 2\sin x\).

So, \(f\left( 0 \right) = 0 + 0 + 2 = 2\), that implies \({f^{ - 1}}\left( 2 \right) = 0\) and \(f'\left( 0 \right) = 0 + 3 - 0 = 3\).

Substitute \(x = 2\) in the above formula:

\(\begin{array}{l}{\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( 2 \right)} \right)}}\\{\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right) = \frac{1}{{f'\left( 0 \right)}}\\{\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right) = \frac{1}{3}\end{array}\)

Hence, \({\left( {{f^{ - 1}}} \right)^\prime }\left( 2 \right) = \frac{1}{3}\).