Question

If $y = {x^n}\log x + x{\left( {\log x} \right)^n}$, then $\dfrac{{dy}}{{dx}}$ is equal to
A.${x^{n - 1}}\left( {1 + n\log x} \right) + {\left( {\log x} \right)^{n - 1}}\left( {n + \log x} \right)$
B.${x^{n - 2}}\left( {1 + n\log x} \right) + {\left( {\log x} \right)^{n - 1}}\left( {n + \log x} \right)$
C.${x^{n - 1}}\left( {1 + n\log x} \right) + {\left( {\log x} \right)^{n - 1}}\left( {n - \log x} \right)$
D.None of these

Answer 1

Hint : In order to find $\dfrac{{dy}}{{dx}}$, differentiate the equation of y with respect to x, and the differentiation can be solved using the product rule, written as $\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$. Substitute the functions of both the operands separately, solve them and get the results.
Formula used:
$\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
$\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d{{\left( {\log x} \right)}^n}}}{{dx}} = \dfrac{1}{x}n{\left( {\log x} \right)^{n - 1}}$
$\dfrac{{d\left( x \right)}}{{dx}} = 1$

Complete step-by-step answer :
We are given with the equation $y = {x^n}\log x + x{\left( {\log x} \right)^n}$, we need to find the value of $\dfrac{{dy}}{{dx}}$.
So, differentiating both the sides of the equation y with respect to x, we get:
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^n}\log x + x{{\left( {\log x} \right)}^n}} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}} + \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}}$. ……..(1)
We can see we have two functions in each operand.
Solving each operand separately.
Taking $\dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}}$:
Since, we have two functions, so we can use product rule of differentiation which is as:
$\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Comparing $uv$ with ${x^n}\log x$, we get:
$u = {x^n} \\\ v = \log x \;$
Substituting these values in the product rule, we get:
$\Rightarrow \dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}} = {x^n}\dfrac{{d\left( {\log x} \right)}}{{dx}} + \log x\dfrac{{d\left( {{x^n}} \right)}}{{dx}}$
Since, we know that $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$ and $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$, substituting them in the above equation.
$\Rightarrow \dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}} = {x^n}\left( {\dfrac{1}{x}} \right) + \log x\left( {n{x^{n - 1}}} \right)$
From the law of radicals, we can write $\dfrac{{{x^n}}}{x} = {x^{n - 1}}$.
$\Rightarrow \dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}} = {x^{n - 1}} + n{x^{n - 1}}\log x$
Taking ${x^{n - 1}}$ common:
$\Rightarrow \dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}} = {x^{n - 1}}\left( {1 + n\log x} \right)$ …….(2)
Similarly, differentiating the other operand $\dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}}$:
Since, we have two functions in this operand also, so we can use product rule of differentiation which is as:
$\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Comparing $uv$ with $x{\left( {\log x} \right)^n}$, we get:
$u = x \\\ v = {\left( {\log x} \right)^n} \\\$
Substituting these values in the product rule, we get:
$\Rightarrow \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}} = x\dfrac{{d{{\left( {\log x} \right)}^n}}}{{dx}} + {\left( {\log x} \right)^n}\dfrac{{d\left( x \right)}}{{dx}}$
Since, we know that $\dfrac{{d{{\left( {\log x} \right)}^n}}}{{dx}} = \dfrac{1}{x}n{\left( {\log x} \right)^{n - 1}}$ and $\dfrac{{d\left( x \right)}}{{dx}} = 1$, substituting them in the above equation.
$\Rightarrow \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}} = x.\dfrac{1}{x}.n{\left( {\log x} \right)^{n - 1}} + {\left( {\log x} \right)^n}$
$\Rightarrow \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}} = n{\left( {\log x} \right)^{n - 1}} + {\left( {\log x} \right)^n}$ ……(3)
Since, we can write ${\left( {\log x} \right)^n}$ as ${\left( {\log x} \right)^{\left( {n - 1} \right) + 1}}$ .
From Law of radicals, we can write ${\left( {\log x} \right)^{\left( {n - 1} \right) + 1}} = \left( {\log x} \right){\left( {\log x} \right)^{\left( {n - 1} \right)}}$ .
Writing it in the equation 3, we get:
$\Rightarrow \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}} = n{\left( {\log x} \right)^{n - 1}} + \left( {\log x} \right){\left( {\log x} \right)^{n - 1}}$
Taking ${\left( {\log x} \right)^{n - 1}}$ common from the above equation:
$\Rightarrow \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}} = {\left( {\log x} \right)^{n - 1}}\left( {n + \log x} \right)$ …….(4)
Substituting the equation 2 and equation 4 in equation 1, we get:
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^n}\log x} \right)}}{{dx}} + \dfrac{{d\left( {x{{\left( {\log x} \right)}^n}} \right)}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = {x^{n - 1}}\left( {1 + n\log x} \right) + {\left( {\log x} \right)^{n - 1}}\left( {n + \log x} \right)$
Which is the same as Option 1.
So, the correct answer is “Option 1”.

Note : We could have directly differentiated the operands in the equation but it is always preferred to solve them separately in order to avoid confusion as they are of the same types and the same product rule is being followed. So, to avoid mistakes, solve them separately.
Law of Radical- If in product, we have the same bases, their powers will be added and for division their powers will be subtracted.