Miscellaneous Exercise Q4 Limits and Derivatives - NCERT Maths Class 11


Question

Find the derivative of the following function (it is to be understood that \(a,\;b,\;c,\;d\) are fixed non-zero constants): \(\left( {ax + b} \right){\left( {cx + d} \right)^2}\)

Text Solution

Let \({\rm{ }}f'\left( x \right) = \left( {ax + b} \right){\left( {cx + d} \right)^2}\)

By Leibnitz product rule,

\[\begin{align}f'\left( x \right) &= \left( {ax + b} \right)\frac{d}{{dx}}{\left( {cx + d} \right)^2} + {\left( {cx + d} \right)^2}\frac{d}{{dx}}\left( {ax + b} \right)\\&= \left( {ax + b} \right)\frac{d}{{dx}}\left( {{c^2}{x^2} + 2cd{x^2} + {d^2}} \right) + {\left( {cx + d} \right)^2}\frac{d}{{dx}}\left( {ax + b} \right)\\&= \left( {ax + b} \right)\left[ {\frac{d}{{dx}}\left( {{c^2}{x^2}} \right) + \frac{d}{{dx}}\left( {2cdx} \right) + \frac{d}{{dx}}\left( {{d^2}} \right)} \right] \\&\quad+ {\left( {cx + d} \right)^2}\left[ {\frac{d}{{dx}}\left( {ax} \right) + \frac{d}{{dx}}\left( b \right)} \right]\\&= \left( {ax + b} \right)\left( {2{c^2}x + 2cd} \right) + {\left( {cx + d} \right)^2}a\\&= 2c\left( {ax + b} \right)\left( {cx + d} \right) + a{\left( {cx + d} \right)^2}\end{align}\]

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