Question

Suppose that \(f\)and \(g\) are differentiable functions and let \(h\left( x \right) = f{\left( x \right)^{g\left( x \right)}}\). Use logarithm differentiation to derive the formula

\(h' = g \cdot {f^{g - 1}} \cdot f' + \left( {\ln f} \right) \cdot {f^g} \cdot g'\)

Short answer

The formula has been derived using logarithm differentiation.

Step-by-step solution

The product rule of differentiation

If a function\(h\left( x \right) = f\left( x \right) \cdot g\left( x \right)\)and\(f\),\(g\)are both differentiable, then the derivative of\(h\left( x \right)\)is as follows:

\(\frac{{dh}}{{dx}} = \frac{d}{{dx}}\left( {f\left( x \right) \cdot g\left( x \right)} \right) = f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

The derivative of the given function

It is given that \(h = {f^g}\).

Take log both the sides:

\(\ln h = g\ln f\)

Now, differentiate both the sides with respect to \(x\) as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {\ln h} \right) &= \frac{d}{{dx}}\left( {g \cdot \ln f} \right)\\\frac{1}{h} \cdot \frac{d}{{dx}}\left( h \right) &= g \cdot \frac{1}{f} \cdot \frac{d}{{dx}}\left( f \right) + \left( {\ln f} \right)\frac{d}{{dx}}\left( g \right)\\h' &= h\left( {g \cdot \frac{1}{f} \cdot f' + \left( {\ln f} \right)g'} \right)\\h' &= {f^g}\left( {g \cdot \frac{1}{f} \cdot f' + \left( {\ln f} \right) \cdot g'} \right)\\h' &= g \cdot {f^{g - 1}} \cdot f' + \left( {\ln f} \right) \cdot {f^g} \cdot g'\end{aligned}\)

Hence, it is proved that \(h' = g \cdot {f^{g - 1}} \cdot f' + \left( {\ln f} \right) \cdot {f^g} \cdot g'\).