Question

If \(f\) and \(g\) are the functions whose graphs are shown, let \(u\left( x \right) = f\left( {g\left( x \right)} \right)\), \(v\left( x \right) = g\left( {f\left( x \right)} \right)\) and \(w\left( x \right) = g\left( {g\left( x \right)} \right)\). Find each derivative, if it exists. If it does not exist, explain why.

(a) \(u'\left( 1 \right)\) (b) \(v'\left( 1 \right)\) (c) \(w'\left( 1 \right)\)

Short answer

(a) The required value is \(u'\left( 1 \right) = \frac{1}{4}\) .

(b) The required value is \(v'\left( 1 \right) = - 2\) .

(c) The required value is \(w'\left( 1 \right) = - \frac{1}{2}\) .

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 \(m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\).

(a) Step 3: The value of \(u'\left( 1 \right)\)

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

Now, from the graphs of \(f\) and \(g\), it is clear that \(g\left( 1 \right) = 4\) because the graph of \(g\) passes through \(\left( {1,4} \right)\).

The slope of line segment between \(\left( {2,4} \right)\) and \(\left( {6,3} \right)\) is obtained below:

\(\begin{aligned}m &= \frac{{3 - 4}}{{6 - 2}}\\ &= \frac{{ - 1}}{4}\end{aligned}\)

So, \(f'\left( 4 \right) = - \frac{1}{4}\) because the derivative of straight-line function is equal to the slope of the line.

The slope of line segment between \(\left( {0,5} \right)\) and \(\left( {3,2} \right)\) is obtained below:

\(\begin{aligned}m &= \frac{{2 - 5}}{{3 - 0}}\\ &= \frac{{ - 3}}{3}\\ &= - 1\end{aligned}\)

So, \(g'\left( 1 \right) = - 1\) because the derivative of straight-line function is equal to the slope of the line.

Now, substitute these values in the formula of \(u'\left( 1 \right)\) and solve as follows:

\(\begin{aligned}u'\left( 1 \right) &= f'\left( {g\left( 1 \right)} \right) \cdot g'\left( 1 \right)\\ &= f'\left( 4 \right) \cdot \left( { - 1} \right)\\ &= - \frac{1}{4} \cdot \left( { - 1} \right)\\ &= \frac{1}{4}\end{aligned}\)

Hence, the value of \(u'\left( 1 \right)\) is \(\frac{1}{4}\).

(b) Step 4: The value of \(v'\left( 1 \right)\)

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

So, \(v'\left( 1 \right) = g'\left( {f\left( 1 \right)} \right) \cdot f'\left( 1 \right)\).

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

The slope of line segment between \(\left( {0,0} \right)\) and \(\left( {2,4} \right)\) is obtained below:

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

So, \(f'\left( 1 \right) = 2\) because the derivative of straight-line function is equal to the slope of the line.

The slope of line segment between \(\left( {0,5} \right)\) and \(\left( {3,2} \right)\) is obtained below:

\(\begin{aligned}m &= \frac{{2 - 5}}{{3 - 0}}\\ &= \frac{{ - 3}}{3}\\ &= - 1\end{aligned}\)

So, \(g'\left( 2 \right) = - 1\) because the derivative of straight-line function is equal to the slope of the line.

Now, substitute these values in the formula of \(v'\left( 1 \right)\) and solve as follows:

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

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

(c) Step 5: The value of \(w'\left( 1 \right)\)

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

So, \(w'\left( 1 \right) = g'\left( {g\left( 1 \right)} \right) \cdot g'\left( 1 \right)\).

Now, from the graphs of \(f\) and \(g\), it is clear that \(g\left( 1 \right) = 4\) because the graph of \(f\) passes through \(\left( {1,4} \right)\).

The slope of line segment between \(\left( {0,5} \right)\) and \(\left( {3,2} \right)\) is obtained below:

\(\begin{aligned}m &= \frac{{2 - 5}}{{3 - 0}}\\ &= \frac{{ - 3}}{3}\\ &= - 1\end{aligned}\)

So, \(g'\left( 2 \right) = - 1\) because the derivative of straight-line function is equal to the slope of the line.

The slope of line segment between \(\left( {3,2} \right)\) and \(\left( {7,4} \right)\) is obtained below:

\(\begin{aligned}m &= \frac{{4 - 2}}{{7 - 3}}\\ &= \frac{2}{4}\\ &= \frac{1}{2}\end{aligned}\)

So, \(g'\left( 4 \right) = \frac{1}{2}\) because the derivative of straight-line function is equal to the slope of the line.

Now, substitute these values in the formula of \(w'\left( 1 \right)\) and solve as follows:

\(\begin{aligned}w'\left( 1 \right) &= g'\left( {g\left( 1 \right)} \right) \cdot g'\left( 1 \right)\\ &= g'\left( 4 \right) \cdot \left( { - 1} \right)\\ &= \frac{1}{2} \cdot \left( { - 1} \right)\\ &= - \frac{1}{2}\end{aligned}\)

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