Question

If \(f\) is a differentiable function, find an expression for the

derivative of each of the following functions.

(a) \(y = {x^2}f\left( x \right)\) (b) \(y = \frac{{f\left( x \right)}}{{{x^2}}}\)

(c) \(y = \frac{{{x^2}}}{{f\left( x \right)}}\) (d) \(y = \frac{{1 + xf\left( x \right)}}{{\sqrt x }}\)

Short answer

(a) The required expression is \(y' = xf'\left( x \right) + 2xf\left( x \right)\).

(b) The required expression is \(y' = \frac{{xf'\left( x \right) - 2f\left( x \right)}}{{{x^3}}}\).

(c) The required expression is \(y' = \frac{{2xf\left( x \right) - {x^2}f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}\).

(d) The required expression is \(y' = \frac{{2{x^2}f'\left( x \right) + xf\left( x \right) - 1}}{{2{x^{\frac{3}{2}}}}}\).

Step-by-step solution

Derivative of division of two function

If a function is division of two function the 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 derivativewill be

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

Finding the derivative of the given function

(a) The function is given by \(y = {x^2}f\left( x \right)\).

Differentiate the function with respect to \(x\) we get,

\(\begin{aligned}y' & = \frac{d}{{dx}}{x^2}f\left( x \right)\\ & = {x^2}f'\left( x \right) + 2xf\left( x \right)\end{aligned}\)

Hence \(y' = xf'\left( x \right) + 2xf\left( x \right)\).

Finding the derivative of the given function

(b) The function is given by \(y = \frac{{f\left( x \right)}}{{{x^2}}}\) .

Differentiate the function with respect to \(x\) we get

\(\begin{aligned}y' & = \frac{{{x^2}\frac{d}{{dx}}f\left( x \right) - f\left( x \right)\frac{d}{{dx}}{x^2}}}{{{x^4}}}\\ & = \frac{{{x^2}f'\left( x \right) - 2xf\left( x \right)}}{{{x^4}}}\\ & = \frac{{xf'\left( x \right) - 2f\left( x \right)}}{{{x^3}}}\end{aligned}\)

Hence\(y' = \frac{{xf'\left( x \right) - 2f\left( x \right)}}{{{x^3}}}\) .

Finding the derivative of the given function

(c) The function is given by \(y = \frac{{{x^2}}}{{f\left( x \right)}}\) .

Differentiate the function with respect to \(x\) we get

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

Hence \(y' = \frac{{2xf\left( x \right) - {x^2}f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}\).

Finding the derivative of the given function

(d) The function is given by \(y = \frac{{1 + xf\left( x \right)}}{{\sqrt x }}\).

Differentiate the function with respect to \(x\) we get,

\(\begin{aligned}y' & = \frac{{\sqrt x \frac{d}{{dx}}\left( {1 + xf\left( x \right)} \right) - \left( {1 + xf\left( x \right)} \right)\frac{d}{{dx}}\sqrt x }}{x}\\ & = \frac{{\sqrt x \left( {xf'\left( x \right) + f\left( x \right)} \right) - \frac{{\left( {1 + xf\left( x \right)} \right)}}{{2\sqrt x }}}}{x}\\ & = \frac{{2x\left( {xf'\left( x \right) + f\left( x \right)} \right) - 1 - xf\left( x \right)}}{{2{x^{\frac{3}{2}}}}}\\ & = \frac{{2{x^2}f'\left( x \right) + xf\left( x \right) - 1}}{{2{x^{\frac{3}{2}}}}}\end{aligned}\)

Hence \(y' = \frac{{2{x^2}f'\left( x \right) + xf\left( x \right) - 1}}{{2{x^{\frac{3}{2}}}}}\).