Question

If \(f\) and \(g\) are the functions whose graphs are shown, let \(u\left( x \right) = f\left( x \right)g\left( x \right)\) and \(v\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\).

(a) Find \(u'\left( 1 \right)\) (b) Find \(v'\left( 4 \right)\)

Short answer

(a) The required value is \(u'\left( 1 \right) = 3\).

(b) The required value is \(v'\left( 4 \right) = - \frac{7}{{12}}\).

Step-by-step solution

Determine the slope of a line segment passing through a point

The slope of a line segment passes through the point \(\left( {c,f\left( c \right)} \right)\) with endpoints \(\left( {a,f\left( a \right)} \right)\) and \(\left( {b,f\left( b \right)} \right)\) is defined by \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\).

Finding the derivative of a function at a point

(a) Given that \(u\left( x \right) = f\left( x \right)g\left( x \right)\).

Then we get by differentiating,

\(u'\left( x \right) = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)\).

Now to find \(u'\left( 1 \right)\) we have to find \(f'\left( 1 \right)\) and \(g'\left( 1 \right)\) first.

Since the line segment with the points \(\left( {1,f\left( 1 \right)} \right)\) and \(\left( {4,f\left( 4 \right)} \right)\) passing through the point \(\left( {1,f\left( 1 \right)} \right)\) so the derivative of \(f\) at the point \(x = 1\) will be,

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

Also, since the line segment with the points \(\left( {0,g\left( 0 \right)} \right)\) and \(\left( {2,g\left( 2 \right)} \right)\) passing through the point \(\left( {1,g\left( 1 \right)} \right)\) so the derivative at the point \(x = 1\) will be,

\(\begin{aligned}g'\left( 1 \right) & = \frac{{g\left( 2 \right) - g\left( 0 \right)}}{2}\\ & = \frac{{4 - 2}}{2}\\ & = 1\end{aligned}\)

Now since \(f\left( 1 \right) = 2\) and \(g\left( 2 \right) = 3\) so derivative of the function \(u\) at the point \(x = 1\) will be,

\(\begin{aligned}u'\left( 1 \right) & = f\left( 1 \right)g'\left( 1 \right) + g\left( 1 \right)f'\left( 1 \right)\\ & = 2 \times 1 + 3 \times \frac{1}{3}\\ & = 2 + 1\\ & = 3\end{aligned}\)

Hence \(u'\left( 1 \right) = 3\).

Finding the derivative of a function at a point

(b) Given that \(v\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\).

Then we get by differentiating,

\(v'\left( x \right) = \frac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\).

Now to find \(v'\left( 4 \right)\) we have to find \(f'\left( 4 \right)\) and \(g'\left( 4 \right)\) first.

Since the line segment with the points \(\left( {1,f\left( 1 \right)} \right)\) and \(\left( {7,f\left( 7 \right)} \right)\) passing through the point \(\left( {4,f\left( 4 \right)} \right)\) so the derivative of \(f\) at the point \(x = 4\) will be ,

\(\begin{aligned}f'\left( 4 \right) & = \frac{{f\left( 7 \right) - f\left( 1 \right)}}{6}\\ & = \frac{{4 - 2}}{6}\\ & = \frac{1}{3}\end{aligned}\)

Since the graph \(g\) line segment passes through the point \(\left( {1,g\left( 1 \right)} \right)\) and \(\left( {4,g\left( 4 \right)} \right)\) is parallel to each other, hence they have the same slope. So \(g'\left( 4 \right) = 1\).

Now since \(f\left( 4 \right) = 3\) and \(g\left( 4 \right) = 2\) so derivative of the function \(v\) at the point \(x = 4\) will be,

\(\begin{aligned}v'\left( 4 \right) & = \frac{{g\left( 4 \right)f'\left( 4 \right) - f\left( 4 \right)g'\left( 4 \right)}}{{{{\left( {g\left( 4 \right)} \right)}^2}}}\\ & = \frac{{2 \times \frac{1}{3} - 3 \times 1}}{4}\\ & = \frac{{\frac{{2 - 9}}{3}}}{4}\\ & = - \frac{7}{{12}}\end{aligned}\)

Hence \(v'\left( 4 \right) = - \frac{7}{{12}}\) .