Question

If \(g\left( x \right) = \sqrt {f\left( x \right)} \), where the graph of \(f\) is shown evaluate \(g'\left( 3 \right)\).

Short answer

The value of \(g'\left( 3 \right)\) is \( - \frac{1}{{3\sqrt 2 }}\).

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

The slope of a line

The slope of a line passing through two points \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right)\) can be obtained using the formula \(m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\).

The value of \(g'\left( 3 \right)\)

It is given that \(g\left( x \right) = \sqrt {f\left( x \right)} \). So, by chain rule \(g'\left( x \right) = \frac{1}{{2\sqrt {f\left( x \right)} }} \cdot f'\left( x \right)\).

So, \(g'\left( 3 \right) = \frac{1}{{2\sqrt {f\left( 3 \right)} }} \cdot f'\left( 3 \right)\).

Now, from the graphs of \(f\), it is clear that \(f\left( 3 \right) = 2\) because the graph of \(f\) passes through \(\left( {3,2} \right)\).

The slope of line segment between \(\left( {0,4} \right)\) and \(\left( {6,0} \right)\) is obtained below:

\(\begin{aligned}m &= \frac{{0 - 4}}{{6 - 0}}\\ &= \frac{{ - 4}}{6}\\ &= - \frac{2}{3}\end{aligned}\)

So, \(f'\left( 3 \right) = - \frac{2}{3}\) because the derivative of straight-line function is equal to the slope of the line.

Now, substitute these values in the formula of \(g'\left( 3 \right)\) and solve as follows:

\(\begin{aligned}g'\left( 3 \right) &= \frac{{f'\left( 3 \right)}}{{2\sqrt {f\left( 3 \right)} }}\\ &= \frac{{ - \frac{2}{3}}}{{2\sqrt 2 }}\\ &= - \frac{1}{{3\sqrt 2 }}\end{aligned}\)

Hence, the value of \(g'\left( 3 \right)\) is \( - \frac{1}{{3\sqrt 2 }}\).