Question

Differentiate.

30. \(f\left( x \right) = \frac{{ax + b}}{{cx + d}}\)

Short answer

The answer is \(f'\left( x \right) = \frac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\).

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

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

Finding the derivative of the given function

The function is given by \(f\left( x \right) = \frac{{ax + b}}{{cx + d}}\).

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

\(\begin{aligned}f'\left( x \right) & = \frac{d}{{dx}}\left( {\frac{{ax + b}}{{cx + d}}} \right)\\ & = \frac{{\left( {cx + d} \right)\frac{d}{{dx}}\left( {ax + b} \right) - \left( {ax + b} \right)\frac{d}{{dx}}\left( {cx + d} \right)}}{{{{\left( {cx + d} \right)}^2}}}\\ & = \frac{{a\left( {cx + d} \right) - c\left( {ax + b} \right)}}{{{{\left( {cx + d} \right)}^2}}}\\ & = \frac{{acx + ad - acx - bc}}{{{{\left( {cx + d} \right)}^2}}}\\ & = \frac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\end{aligned}\)

Hence \(f'\left( x \right) = \frac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\).