Question

Let \(f\) and \(g\) be the functions in Exercise 69.

(a) If \(F\left( x \right) = f\left( {f\left( x \right)} \right)\), find \(F'\left( 2 \right)\).

(b) If \(G\left( x \right) = g\left( {g\left( x \right)} \right)\), find \(G'\left( 3 \right)\).

Short answer

(a) The value of \(F'\left( 2 \right)\) is \(20\).

(b) The value of \(G'\left( 3 \right)\) is \(63\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

(a) Step 2: The value of \(F'\left( 2 \right)\)

It is given that \(F\left( x \right) = f\left( {f\left( x \right)} \right)\), then by chain rule \(F'\left( x \right) = f'\left( {f\left( x \right)} \right) \cdot f'\left( x \right)\).

Put \(x = 2\) in the above formula and solve as follows:

\(\begin{aligned}F'\left( 2 \right) &= f'\left( {f\left( 2 \right)} \right) \cdot f'\left( 2 \right)\\ &= f'\left( 1 \right) \cdot 5\\ &= 4 \cdot 5\\ &= 20\end{aligned}\)

Hence, the value of \(F'\left( 2 \right)\) is \(20\).

(b) Step 3: The value of \(G'\left( 3 \right)\)

It is given that \(G\left( x \right) = g\left( {g\left( x \right)} \right)\), then by chain rule \(G'\left( x \right) = g'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Put \(x = 3\) in the above formula and solve as follows:

\(\begin{aligned}G'\left( 3 \right) &= g'\left( {g\left( 3 \right)} \right) \cdot g'\left( 3 \right)\\ &= g'\left( 2 \right) \cdot 9\\ &= 7 \cdot 9\\ &= 63\end{aligned}\)

Hence, the value of \(G'\left( 3 \right)\) is \(63\).