Question

Differentiate w.r.t. x in $nx{\left( {1 + x} \right)^{n - 1}}$

Answer 1

Use the product rule to differentiate the given function where product rule is used to find the derivatives of the product of two or more functions. If u and v are the given function of x, then their product rule formula is given by
$\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
From the given formula, we can see the first function is multiplied with the derivative of the second function, and the derivative of the first function is multiplied with the second function.

In this question, we can see there are two functions whose differentiation is to be found; hence we use the product rule.
Basic formulae of derivation to be used in this question are:

$$\dfrac{d}{{dx}}\left( c \right) = 0 \\\ \dfrac{d}{{dx}}\left( x \right) = 1 \\\ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\\ \dfrac{{d\left( {{a^x}} \right)}}{{dx}} = {a^{^x}}\ln a \\\$$

Complete step by step solution:
Let, $y = nx{\left( {1 + x} \right)^{n - 1}} - - (i)$
Here the given function contains two functions; hence we use product rule where
$u = x$ and $v = {\left( {1 + x} \right)^{n - 1}}$
So by using product rule to differentiate equation (i), we get

\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {nx{{\left( {1 + x} \right)}^{n - 1}}} \right] \\\ = n{\left( {1 + x} \right)^{n - 1}} + nx\left( {n - 1} \right){\left( {1 + x} \right)^{n - 2}} \\\ = n{\left( {1 + x} \right)^{n - 2}}\left[ {\left( {1 + x} \right) + x\left( {n - 1} \right)} \right] \\\ $$[Since$$\dfrac{d}{{dx}}\left( x \right) = 1$$] By further solving the obtained equation, we can write

\dfrac{{dy}}{{dx}} = n{\left( {1 + x} \right)^{n - 2}}\left[ {1 + x + nx - x} \right] \\
\dfrac{{dy}}{{dx}} = n{\left( {1 + x} \right)^{n - 2}}\left[ {1 + nx} \right] \\

**Hence we can say differentiation of$$nx{\left( {1 + x} \right)^{n - 1}}$$ w.r.t. x $$ = n\left( {1 + nx} \right){\left( {1 + x} \right)^{n - 2}}$$** **Note:** One of the special rules of the product rule is the constant multiple rules, where c is a number, and f(x) is a differentiable function then $$cf\left( x \right)$$ is also differentiable, and its derivative is $${\left( {cf} \right)^\prime } = cf'\left( x \right)$$, where differentiation of a constant number is equal to zero.