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

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

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)\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=(ax+b)n(cx+d)my = {\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m}
Now differentiate it w.r.t. x we have,
⇒ddxy=ddx[(ax+b)n(cx+d)m]\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
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)nddx(cx+d)m+(cx+d)mddx(ax+b)n\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 ddx(cx+d)m=m(cx+d)m−1ddx(cx+d)\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,
⇒ddxy=(ax+b)nm(cx+d)m−1ddx(cx+d)+(cx+d)mn(ax+b)n−1ddx(ax+b)\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,
⇒ddxy=(ax+b)nm(cx+d)m−1(c+0)+(cx+d)mn(ax+b)n−1(a+0)\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,
⇒ddxy=cm(ax+b)n(cx+d)m−1+an(cx+d)m(ax+b)n−1\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}}
⇒ddxy=cm(ax+b)n(cx+d)m(cx+d)+an(cx+d)m(ax+b)n(ax+b)\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 (ax+b)n(cx+d)m{\left( {ax + b} \right)^n}{\left( {cx + d} \right)^m} common we have,
⇒ddxy=(ax+b)n(cx+d)m[cm(cx+d)+an(ax+b)]\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 (cx+d)m{\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 (cx+d)m{\left( {cx + d} \right)^m} as above and simplify we will get the required answer.