Question

If \[y = \dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}\] , then find \[\dfrac{{dy}}{{dx}}\] .
A.\[\left[ {\dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}} \right]\left[ {\left( {\dfrac{3}{2}} \right)\dfrac{{\left( {1 - \cos x} \right)}}{{\left( {1 - \sin x} \right)}} - \dfrac{1}{{2x}}} \right]\]
B. \[\left[ {\dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}} \right]\left[ {\left( {\dfrac{3}{2}} \right)\dfrac{{\left( {1 - \cos x} \right)}}{{\left( {x - \sin x} \right)}} - \dfrac{1}{{2x}}} \right]\]
C. \[\left[ {\dfrac{{2{{(x - \sin x)}^{\dfrac{1}{2}}}}}{{\sqrt x }}} \right]\left[ {\left( {\dfrac{3}{2}} \right)\dfrac{{\left( {1 - \cos x} \right)}}{{\left( {1 - \sin x} \right)}} - \dfrac{1}{{2x}}} \right]\]
D. None of these

Answer 1

Hint: Take logarithm to both sides of the given equation. Use some formulas of logarithm to simplify the obtained expression. Now, differentiate both sides of the obtained equation with respect to x to obtain the required answer.

Formula used:
\[\log MN = \log M + \log N\]
\[\dfrac{{\log M}}{{\log N}} = \log M - \log N\]
\[\log {a^x} = x\log a\]
The chain rule of derivative is,
\[\dfrac{d}{{dx}}f(g(x)) = \dfrac{{df}}{{dg(x)}}.\dfrac{{dg(x)}}{{dx}}\]
\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]
\[\dfrac{d}{{dx}}(x) = 1\]
\[\dfrac{d}{{dx}}(\sin x) = \cos x\]

Complete step by step solution:
The given equation is,
\[y = \dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}\]
Take logarithm to both sides of the given equation \[y = \dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}\],
\[\log y = \log \dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}\]
Apply the stated formulas of logarithm for simplification,
\[\log y = \log 2{(x - \sin x)^{\dfrac{3}{2}}} - \log \sqrt x \]
\[ = \log 2 + \log {(x - \sin x)^{\dfrac{3}{2}}} - \log {x^{\dfrac{1}{2}}}\]
\[ = \log 2 + \dfrac{3}{2}\log (x - \sin x) - \dfrac{1}{2}\log x\]
Differentiate the equation \[\log y = \log 2 + \dfrac{3}{2}\log (x - \sin x) - \dfrac{1}{2}\log x\] with respect to x.
\[\dfrac{1}{y}\dfrac{{dy}}{{dx}} = 0 + \dfrac{3}{2}\dfrac{1}{{(x - \sin x)}}\dfrac{d}{{dx}}(x - \sin x) - \dfrac{1}{2}.\dfrac{1}{x}\]
\[ = \dfrac{3}{2}\dfrac{1}{{(x - \sin x)}}(1 - \cos x) - \dfrac{1}{{2x}}\]
Multiply y to both sides of the equation \[\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{3}{2}\dfrac{1}{{(x - \sin x)}}(1 - \cos x) - \dfrac{1}{{2x}}\]to obtain the required result.
\[\dfrac{{dy}}{{dx}} = y\left[ {\dfrac{3}{2}\dfrac{{(1 - \cos x)}}{{(x - \sin x)}} - \dfrac{1}{{2x}}} \right]\]
Now, substitute the given value of y in the equation \[\dfrac{{dy}}{{dx}} = y\left[ {\dfrac{3}{2}\dfrac{{(1 - \cos x)}}{{(x - \sin x)}} - \dfrac{1}{{2x}}} \right]\].
\[\dfrac{{dy}}{{dx}} = \dfrac{{2{{(x - \sin x)}^{\dfrac{3}{2}}}}}{{\sqrt x }}\left[ {\dfrac{3}{2}\dfrac{{(1 - \cos x)}}{{(x - \sin x)}} - \dfrac{1}{{2x}}} \right]\]

The correct option is B.

Note: When we are asked to find the derivation of the given equation, we need to change the given equation for convenience. Students make mistakes if they don’t use logarithm formulas to modify the given equation and it leads to a complex expression.