Question

If \(F\left( x \right) = f\left( {3f\left( {4f\left( x \right)} \right)} \right)\), where \(f\left( 0 \right) = 0\) and \(f'\left( 0 \right) = 2\), find \(F'\left( 0 \right)\).

Short answer

The value of \(F'\left( 0 \right)\) is 96.

Step-by-step solution

The Chain Rule

The composite function \(F = f \circ g\), or \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiableat \(x\) and \(F'\) is obtained as:

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

In Leibniz notation, the derivative is shown below:

\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

Determine the value of \(F'\left( 0 \right)\)

Use the chain rule to differentiate the function \(F\left( x \right)\) as shown below:

\(\begin{aligned}F'\left( x \right) &= \frac{d}{{dx}}f\left( {3f\left( {4f\left( x \right)} \right)} \right)\\ &= f'\left( {3f\left( {4f\left( x \right)} \right)} \right)\frac{d}{{dx}}\left( {3f\left( {4f\left( x \right)} \right)} \right)\\ &= f'\left( {3f\left( {4f\left( x \right)} \right)} \right) \cdot 3f'\left( {4f\left( x \right)} \right)\frac{d}{{dx}}\left( {4f\left( x \right)} \right)\\ &= f'\left( {3f\left( {4f\left( x \right)} \right)} \right) \cdot 3f'\left( {4f\left( x \right)} \right) \cdot 4f'\left( x \right)\end{aligned}\)

Use the values of \(f\left( 0 \right),f'\left( 0 \right)\) to determine the value of \(F'\left( 0 \right)\) as shown below:

\(\begin{aligned}F'\left( 0 \right) &= f'\left( {3f\left( {4f\left( 0 \right)} \right)} \right) \cdot 3f'\left( {4f\left( 0 \right)} \right) \cdot 4f'\left( 0 \right)\\ &= f'\left( {3f\left( {4 \cdot 0} \right)} \right) \cdot 3f'\left( {4 \cdot 0} \right) \cdot 4 \cdot 2\\ &= f'\left( {3f\left( 0 \right)} \right) \cdot 3f'\left( 0 \right) \cdot 8\\ &= f'\left( {3 \cdot 0} \right) \cdot 3\left( 2 \right) \cdot 8\\ &= f'\left( 0 \right) \cdot 6 \cdot 8\\ &= 2 \cdot 48\\ &= 96\end{aligned}\)

Thus, the value of \(F'\left( 0 \right)\) is 96.