Question

If \(h\left( 2 \right) = 4\)and \(h'\left( 2 \right) = - 3\), find

\({\left. {\frac{d}{{dx}}\left( {\frac{{h\left( x \right)}}{x}} \right)} \right|_{x = 2}}\)

Short answer

The answer is \({\left. {\frac{d}{{dx}}\left( {\frac{{h\left( x \right)}}{x}} \right)} \right|_{x = 2}} = - 2.5\).

Step-by-step solution

Finding derivative when two functions are multiplied

If a function is division of two functions, then we can differentiate the function by the following rule:

Let the function be \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) then the derivative will be

\(\begin{aligned}h'\left( x \right) & = \frac{d}{{dx}}\frac{{f\left( x \right)}}{{g\left( x \right)}}\\ & = \frac{{g\left( x \right)\frac{d}{{dx}}f\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{aligned}\)

Finding the derivative of the function at a given point

Given that \(f\left( x \right) = \frac{{h\left( x \right)}}{x}\).

Differentiating the function, we get

\(\begin{aligned}f'\left( x \right) & = \frac{d}{{dx}}\left( {\frac{{h\left( x \right)}}{x}} \right)\\ & = \frac{{x\frac{d}{{dx}}h\left( x \right) - h\left( x \right)\frac{d}{{dx}}x}}{{{x^2}}}\\ & = \frac{{xh'\left( x \right) - h\left( x \right)}}{{{x^2}}}\end{aligned}\)

Substitute 2 for x into the function of \(f'\left( x \right)\) to determine \(f'\left( 2 \right)\), that is, \({\left. {\frac{d}{{dx}}\left( {\frac{{h\left( x \right)}}{x}} \right)} \right|_{x = 2}}\).

\(\begin{aligned}f'\left( 2 \right) & = \frac{{2h'\left( 2 \right) - h\left( 2 \right)}}{4}\\{\left. {\frac{d}{{dx}}\left( {\frac{{h\left( x \right)}}{x}} \right)} \right|_{x = 2}} & = \frac{{2 \times \left( { - 3} \right) - 4}}{4}\\ & = \frac{{ - 10}}{4}\\ & = - 2.5\end{aligned}\).

Therefore, \({\left. {\frac{d}{{dx}}\left( {\frac{{h\left( x \right)}}{x}} \right)} \right|_{x = 2}} = - 2.5\).