Question

If \(F = f \circ g\), where f and t are twice differentiable functions,

use the Chain Rule and the Product Rule to show that the second derivative of \(F\) is given by

\(F''\left( x \right) = f''\left( {g\left( x \right)} \right) \cdot {\left( {g'\left( x \right)} \right)^2} + f'\left( {g\left( x \right)} \right) \cdot g''\left( x \right)\)

Short answer

It is proved that \(F''\left( x \right) = f''\left( {g\left( x \right)} \right) \cdot {\left( {g'\left( x \right)} \right)^2} + f'\left( {g\left( x \right)} \right) \cdot g''\left( x \right)\).

Step-by-step solution

Chain Rule of Derivative

Let \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) be a composition of function. The derivative of this function with respect to \(x\) is

\(\begin{aligned}F'\left( x \right) &= \frac{d}{{dx}}f\left( {g\left( x \right)} \right)\\ &= f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\end{aligned}\)

So \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Alternate proof of Quotient Rule

Given \(F\left( x \right) = f\left( {g\left( x \right)} \right)\). Differentiating we get:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Again differentiating we get:

\(\begin{aligned}F''\left( x \right) &= \frac{d}{{dx}}f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\\ &= g'\left( x \right)\frac{d}{{dx}}f'\left( {g\left( x \right)} \right) + f'\left( {g\left( x \right)} \right)\frac{d}{{dx}}g'\left( x \right)\\ &= g'\left( x \right)f''\left( {g\left( x \right)} \right)g'\left( x \right) + f'\left( {g\left( x \right)} \right) \cdot g''\left( x \right)\\ &= f''\left( {g\left( x \right)} \right) \cdot {\left( {g'\left( x \right)} \right)^2} + f'\left( {g\left( x \right)} \right) \cdot g''\left( x \right)\end{aligned}\)

Hence proved.