Question

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

use the Chain Rule to show that

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

Short answer

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

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:

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

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

Alternate proof of Quotient Rule

Let \(p\left( x \right) = g\left( {h\left( x \right)} \right)\). Then we get \(F\left( x \right) = f\left( {p\left( x \right)} \right)\).

Now differentiating the function with respect to \(x\)we get:

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

Also \(p'\left( x \right) = g'\left( {h\left( x \right)} \right) \cdot h'\left( x \right)\).

Finally we get \(F'\left( x \right) = f'\left( {g\left( {h\left( x \right)} \right)} \right) \cdot g'\left( {h\left( x \right)} \right) \cdot h'\left( x \right)\).

Hence it is proved.