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

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

Explanation

Solution

Hint: In this question apply the product rule of differentiation which is given as ddx(uv)=uddxv+vddxu\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 (cx+d)m{\left( {cx + d} \right)^m} which is given as ddx(cx+d)m=m(cx+d)m−1ddx(cx+d) and ddx(ax+b)=a\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=(ax+b)(cx+d)2y = \left( {ax + b} \right){\left( {cx + d} \right)^2}
Now differentiate it w.r.t. x we have,
⇒ddxy=ddx[(ax+b)(cx+d)2]\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
ddx(uv)=uddxv+vddxu\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,
⇒ddxy=(ax+b)ddx(cx+d)2+(cx+d)2ddx(ax+b)\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 ddx(cx+d)2=2(cx+d)ddx(cx+d)\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,
⇒ddxy=(ax+b)2(cx+d)ddx(cx+d)+(cx+d)2(a+0)\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,
⇒ddxy=(ax+b)2(cx+d)(c+0)+(cx+d)2(a+0)\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,
⇒ddxy=(cx+d)[2c(ax+b)+a(cx+d)]\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 (cx+d)m{\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 (cx+d)m{\left( {cx + d} \right)^m}, (ax + b) and constant terms as above and simplify we will get the required answer.