Question

If $y = \dfrac{{2{{\left( {x - \sin x} \right)}^{\dfrac{3}{2}}}}}{{\sqrt x }}$ then find $\dfrac{{dy}}{{dx}}$.
(A) $\dfrac{y}{2}\left[ {3 \cdot \dfrac{{1 - \cos x}}{{x - \sin x}} - \dfrac{1}{x}} \right]$
(B) $\dfrac{y}{2}\left[ {2 \cdot \dfrac{{1 - \cos x}}{{x - \sin x}} - \dfrac{1}{x}} \right]$
(C) $\dfrac{y}{2}\left[ {\dfrac{{1 - \cos x}}{{x - \sin x}} - \dfrac{3}{x}} \right]$
(D) $\dfrac{y}{2}\left[ {\dfrac{{1 - \cos x}}{{x - \sin x}} - \dfrac{2}{x}} \right]$

Answer 1

Hint : In this problem, to find the required derivative first we will take logarithm on both sides of the given function $y = f\left( x \right)$. Then, we will use basic properties of logarithm. Then, we will find the derivative of each term by using a formula of differentiation or chain rule.

Complete step-by-step answer :
In this problem, it is given that $y = \dfrac{{2{{\left( {x - \sin x} \right)}^{\dfrac{3}{2}}}}}{{\sqrt x }} \cdots \cdots \left( 1 \right)$. Let us take a logarithm on both sides of the equation $\left( 1 \right)$. Note that $\sqrt x$ can be written as ${x^{\dfrac{1}{2}}}$. So, we can write
$\log \left( y \right) = \log \left[ {\dfrac{{2{{\left( {x - \sin x} \right)}^{\dfrac{3}{2}}}}}{{\sqrt x }}} \right]\, \cdots \cdots \left( 2 \right)$
Let us use the property $\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$ on the RHS of equation $\left( 2 \right)$. So, we get
$\log y = \log \left[ {2{{\left( {x - \sin x} \right)}^{\dfrac{3}{2}}}} \right] - \log \left( {{x^{\dfrac{1}{2}}}} \right) \cdots \cdots \left( 3 \right)$
Let us use the properties $\log \left( {ab} \right) = \log a + \log b$ and $\log {a^b} = b\log a$ on RHS of equation $\left( 3 \right)$. So, we get
$\log y = \log 2 + \log {\left( {x - \sin x} \right)^{\dfrac{3}{2}}} - \dfrac{1}{2}\log x \\\ \Rightarrow \log y = \log 2 + \dfrac{3}{2}\log \left( {x - \sin x} \right) - \dfrac{1}{2}\log x \cdots \cdots \left( 4 \right) \\\$
Now we are going to differentiate on both sides of equation $\left( 4 \right)$ with respect to $x$. So, we get
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = 0 + \dfrac{3}{2}\left( {\dfrac{1}{{x - \sin x}}} \right)\dfrac{d}{{dx}}\left( {x - \sin x} \right) - \dfrac{1}{2}\left( {\dfrac{1}{x}} \right) \\\ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{3}{2}\left( {\dfrac{1}{{x - \sin x}}} \right)\left( {1 - \cos x} \right) - \dfrac{1}{{2x}} \cdots \cdots \left( 5 \right) \\\$
Note that here we used the following formulas of differentiation. Also we used chain rule.
$\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x} \\\ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\\ \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x \\\$
Also note that $\log 2$ is a constant term. So, its differentiation is zero. Let us simplify the equation $\left( 5 \right)$ to find $\dfrac{{dy}}{{dx}}$. So, we get
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left[ {\dfrac{{3\left( {1 - \cos x} \right)}}{{x - \sin x}} - \dfrac{1}{x}} \right] \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{2}\left[ {3 \cdot \dfrac{{1 - \cos x}}{{x - \sin x}} - \dfrac{1}{x}} \right] \\\$
So, the correct answer is “Option A”.

Note : In this problem, also we can find $\dfrac{{dy}}{{dx}}$ by using division rule of derivatives which is given by $\dfrac{d}{{dx}}\left[ {\dfrac{u}{v}} \right] = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$. If we use this rule then we have to take $u = 2{\left( {x - \sin x} \right)^{\dfrac{3}{2}}}$ and $v = \sqrt x$.