Question

Differentiate.

29. \(f\left( x \right) = \frac{x}{{x + \frac{c}{x}}}\)

Short answer

The answer is \(f'\left( x \right) = \frac{{2cx}}{{{{\left( {{x^2} + c} \right)}^2}}}\)

Step-by-step solution

Derivative of multiplication 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

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

Finding the derivative of the given function

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

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

\(\begin{aligned}f'\left( x \right) & = \frac{d}{{dx}}\left( {\frac{x}{{x + \frac{c}{x}}}} \right)\\ & = \frac{d}{{dx}}\left( {\frac{{{x^2}}}{{{x^2} + c}}} \right)\\ & = \frac{{\left( {{x^2} + c} \right)\frac{d}{{dx}}\left( {{x^2}} \right) - {x^2}\frac{d}{{dx}}\left( {{x^2} + c} \right)}}{{{{\left( {{x^2} + c} \right)}^2}}}\\ & = \frac{{2x\left( {{x^2} + c} \right) - {x^2} \cdot 2x}}{{{{\left( {{x^2} + c} \right)}^2}}}\\ & = \frac{{2{x^3} + 2cx - 2{x^3}}}{{{{\left( {{x^2} + c} \right)}^2}}}\\ & = \frac{{2cx}}{{{{\left( {{x^2} + c} \right)}^2}}}\end{aligned}\)

Hence \(f'\left( x \right) = \frac{{2cx}}{{{{\left( {{x^2} + c} \right)}^2}}}\).