Question

Find the derivative of the following function:
$\left( {ax + b} \right){\left( {cx + d} \right)^2}$

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 $\dfrac{d}{{dx}}{\left( {cx + d} \right)^m} = m{\left( {cx + d} \right)^{m - 1}}\dfrac{d}{{dx}}\left( {cx + d} \right){\text{ and }}\dfrac{d}{{dx}}\left( {ax + b} \right) = a$ and differentiation of constant terms is zero so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let
$y = \left( {ax + b} \right){\left( {cx + d} \right)^2}$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {\left( {ax + b} \right){{\left( {cx + d} \right)}^2}} \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)\dfrac{d}{{dx}}{\left( {cx + d} \right)^2} + {\left( {cx + d} \right)^2}\dfrac{d}{{dx}}\left( {ax + b} \right)$
Now as we know differentiation of $\dfrac{d}{{dx}}{\left( {cx + d} \right)^2} = 2\left( {cx + d} \right)\dfrac{d}{{dx}}\left( {cx + d} \right)$ so use this property and differentiation of constant term is zero so we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {ax + b} \right)2\left( {cx + d} \right)\dfrac{d}{{dx}}\left( {cx + d} \right) + {\left( {cx + d} \right)^2}\left( {a + 0} \right)$
Now again differentiate (cx + d) we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {ax + b} \right)2\left( {cx + d} \right)\left( {c + 0} \right) + {\left( {cx + d} \right)^2}\left( {a + 0} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {cx + d} \right)\left[ {2c\left( {ax + b} \right) + a\left( {cx + d} \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 ${\left( {cx + d} \right)^m}$ differentiation which is stated above then first apply the product rule as above then use the property of differentiation of ${\left( {cx + d} \right)^m}$, (ax + b) and constant terms as above and simplify we will get the required answer.