Question

Let \(r\left( x \right) = f\left( {g\left( {h\left( x \right)} \right)} \right)\), where \(h\left( 1 \right) = 2\), \(g\left( 2 \right) = 3\), \(h'\left( 1 \right) = 4\), \(g'\left( 2 \right) = 5\), and \(f'\left( 3 \right) = 6\). Find \(r'\left( 1 \right)\).

Short answer

The value of \(r'\left( 1 \right)\) is 120.

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 differentiable at \(x\) 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 \(r'\left( 1 \right)\)

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

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

Use the given values to evaluate the values of \(r'\left( 1 \right)\) as shown below:

\(\begin{aligned}r'\left( 1 \right) &= f'\left( {g\left( {h\left( 1 \right)} \right)} \right) \cdot g'\left( {h\left( 1 \right)} \right) \cdot h'\left( 1 \right)\\ &= f'\left( {g\left( 2 \right)} \right) \cdot g'\left( 2 \right) \cdot 4\\ &= f'\left( 3 \right) \cdot 5 \cdot 4\\ &= 6 \cdot 5 \cdot 4\\ &= 120\end{aligned}\)

Thus, the value of \(r'\left( 1 \right)\) is 120.