Question

Find the derivative of the following function:
${\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m}$

Answer 1

Hint: In this question apply the product rule of differentiation which is given as $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ later on in the solution apply the general differentiation property of ${\left( {cx + d} \right)^m}$ which is given as $\dfrac{d}{{dx}}{\left( {cx + d} \right)^m} = m{\left( {cx + d} \right)^{m - 1}}\dfrac{d}{{dx}}\left( {cx + d} \right)$ and differentiation of constant terms is zero so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let
$y = {\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m}$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{{\left( {ax + b} \right)}^n}{{\left( {cx + d} \right)}^m}} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {\left( {ax + b} \right)^n}\dfrac{d}{{dx}}{\left( {cx + d} \right)^m} + {\left( {cx + d} \right)^m}\dfrac{d}{{dx}}{\left( {ax + b} \right)^n}$
Now as we know differentiation of $\dfrac{d}{{dx}}{\left( {cx + d} \right)^m} = m{\left( {cx + d} \right)^{m - 1}}\dfrac{d}{{dx}}\left( {cx + d} \right)$ so use this property and differentiation of constant term is zero so we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {\left( {ax + b} \right)^n}m{\left( {cx + d} \right)^{m - 1}}\dfrac{d}{{dx}}\left( {cx + d} \right) + {\left( {cx + d} \right)^m}n{\left( {ax + b} \right)^{n - 1}}\dfrac{d}{{dx}}\left( {ax + b} \right)$
Now again differentiate (cx + d) and (ax + b) we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {\left( {ax + b} \right)^n}m{\left( {cx + d} \right)^{m - 1}}\left( {c + 0} \right) + {\left( {cx + d} \right)^m}n{\left( {ax + b} \right)^{n - 1}}\left( {a + 0} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{d}{{dx}}y = cm{\left( {ax + b} \right)^n}{\left( {cx + d} \right)^{m - 1}} + an{\left( {cx + d} \right)^m}{\left( {ax + b} \right)^{n - 1}}$
$ \Rightarrow \dfrac{d}{{dx}}y = cm{\left( {ax + b} \right)^n}\dfrac{{{{\left( {cx + d} \right)}^m}}}{{\left( {cx + d} \right)}} + an{\left( {cx + d} \right)^m}\dfrac{{{{\left( {ax + b} \right)}^n}}}{{\left( {ax + b} \right)}}$
Now take ${\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m}$ common we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m}\left[ {\dfrac{{cm}}{{\left( {cx + d} \right)}} + \dfrac{{an}}{{\left( {ax + b} \right)}}} \right]$
So this is the required differentiation.

Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of ${\left( {cx + d} \right)^m}$ and constant term differentiation which is stated above then first apply the product rule as above then use the property of differentiation of ${\left( {cx + d} \right)^m}$ as above and simplify we will get the required answer.