Question

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

Answer 1

Hint: Differentiate the given function \[y = {(x{\cot ^3}x)^{\dfrac{3}{2}}}\] with respect to x with the help of chain rule and product rule to obtain the required value.

Formula used:
1. \[\dfrac{d}{{dx}}(x) = 1\]
2. \[\dfrac{{d(\cot x)}}{{dx}} = \cos e{c^2}x\]
3. \[\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}\]
4. The product rule of derivative is,
\[\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] .
5. The chain rule of derivative is,
\[\dfrac{d}{{dx}}(f(g(x))=f^{‘}(g(x)).g^{‘}(x)\]

Complete step by step solution:
Suppose that \[x{\cot ^3}x\] is x and differentiate the term \[{(x{\cot ^3}x)^{\dfrac{3}{2}}}\] as \[{(x)^{\dfrac{3}{2}}}\], then differentiate \[x{\cot ^3}x\] with respect to x with the help of product rule. This is known as chain rule.
\[\begin{array}{l}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{(x{\cot ^3}x)^{\dfrac{3}{2}}}\\\end{array}\]
=\[\dfrac{3}{2}{\left( {x{{\cot }^3}x} \right)^{\dfrac{1}{2}}}\dfrac{d}{{dx}}(x{\cot ^3}x)\]
=\[\dfrac{3}{2}{\left( {x{{\cot }^3}x} \right)^{\dfrac{1}{2}}}\left[ {x\dfrac{d}{{dx}}\left( {{{\cot }^3}x} \right) + {{\cot }^3}x\dfrac{d}{{dx}}(x)} \right]\]
Again, suppose that \[{\cot ^3}x\] is x and differentiate the term \[({\cot ^3}x)\]as \[{x^3}\], after that differentiate \[\cot x\] with respect to x.
=\[\dfrac{3}{2}{\left( {x{{\cot }^3}x} \right)^{\dfrac{1}{2}}}\left[ {x.3{{\cot }^2}x\dfrac{d}{{dx}}(\cot x) + {{\cot }^3}x} \right]\]
=\[\dfrac{3}{2}{\left( {x{{\cot }^3}x} \right)^{\dfrac{1}{2}}}\left[ {x.3{{\cot }^2}x( - \cos e{c^2}x) + {{\cot }^3}x} \right]\]
=\[\left( {\dfrac{3}{2}} \right){\left( {x{{\cot }^3}x} \right)^{\dfrac{1}{2}}}\left[ {{{\cot }^3}x - 3x{{\cot }^2}x\cos e{c^2}x} \right]\]

The correct option is A.

Notes: Sometime students get confused with the use of chain rule and forget to differentiate \[\cot x\] after the step \[\dfrac{3}{2}{\left( {x{{\cot }^3}x} \right)^{\dfrac{1}{2}}}\left[ {x\dfrac{d}{{dx}}\left( {{{\cot }^3}x} \right) + {{\cot }^3}x\dfrac{d}{{dx}}(x)} \right]\].