Question

If \(f\) is the function whose graph is shown, let \(h\left( x \right) = f\left( {f\left( x \right)} \right)\) and \(g\left( x \right) = f\left( {{x^2}} \right)\). Use the graph of \(f\) to estimate the value of each derivative.

(a)\(h'\left( 2 \right)\) (b) \(g'\left( 2 \right)\)

Short answer

(a) The required value is \(h'\left( 2 \right) = 1\).

(b) The required value is \(g'\left( 2 \right) = 8\).

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 \(h'\left( 2 \right)\)

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

So, \(h'\left( 2 \right) = f'\left( {f\left( 2 \right)} \right) \cdot f'\left( 2 \right)\).

Now, from the graphs of \(f\), it is clear that \(f\left( 2 \right) = 1\) because the graph of \(f\) passes through \(\left( {2,1} \right)\).

From the graph, it can be noticed that at \(x = 1\) and \(x = 2\), the rise and run of the graph are equal and the function is decreasing. Therefore, the slope of the tangent line to the curve at \(x = 1\) and \(x = 2\) is \( - 1\). So, \(f'\left( 1 \right) = f'\left( 2 \right) = - 1\).

Now, substitute these values in the formula of \(h'\left( 2 \right)\) and solve as follows:

\(\begin{aligned}h'\left( 2 \right) &= f'\left( {f\left( 2 \right)} \right) \cdot f'\left( 2 \right)\\ &= f'\left( 1 \right) \cdot f'\left( 2 \right)\\ &= \left( { - 1} \right) \cdot \left( { - 1} \right)\\ &= 1\end{aligned}\)

Hence, the value of \(h'\left( 2 \right)\) is \(1\).

(b) Step 3: The value of \(h'\left( 2 \right)\)

It is given that \(g\left( x \right) = f\left( {{x^2}} \right)\). So, by chain rule \(g'\left( x \right) = f'\left( {{x^2}} \right) \cdot \frac{d}{{dx}}\left( {{x^2}} \right) = f'\left( {{x^2}} \right) \cdot 2x\).

So,

\(\begin{aligned}g'\left( 2 \right) &= f'\left( {{{\left( 2 \right)}^2}} \right) \cdot {\left( 2 \right)^2}\\ &= 4f'\left( 4 \right)\end{aligned}\)

From the graph, it can be noticed that at \(x = 4\), the rise and run of the graph are in the ratio of \(2:1\) and the function is increasing. Therefore, the slope of the tangent line to the curve at \(x = 4\) is \(2\). So, \(f'\left( 4 \right) = 2\).

Now, substitute these values in the formula of \(g'\left( 2 \right)\) and solve as follows:

\(\begin{aligned}g'\left( 2 \right) &= 4f'\left( 4 \right)\\ &= 4 \cdot 2\\ &= 8\end{aligned}\)

Hence, the value of \(g'\left( 2 \right)\) is \(8\).