# If y =$\dfrac{1}{{\sqrt[3]{{{\text{cosec x + cot x}}}}}}$, find $\dfrac{{{\text{dx}}}}{{{\text{dy}}}}$ .

Answer

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Hint: Look into the table of derivatives of trigonometric functions for cosec x and cot x. Convert the root into power and then differentiate it.

Complete step-by-step answer:

Given Data,

y =$\dfrac{1}{{\sqrt[3]{{{\text{cosec x + cot x}}}}}}$

Transform y such that there is no cube root in the equation, for the ease of solving

$\Rightarrow$ y = $\dfrac{1}{{{{\left( {\cos {\text{ec x + cot x}}} \right)}^{\dfrac{1}{3}}}}} = {\left( {\cos {\text{ec x + cot x}}} \right)^{ - \dfrac{1}{3}}}$

Differentiating y with respect to x

$\Rightarrow$ $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{1}{3}}}$

For a function f = (x + 1)$^2$, $\dfrac{{{\text{df}}}}{{{\text{dx}}}}{\text{ becomes 2(x + 1}}{{\text{)}}^{2 - 1}}\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{x + 1)}}$

Similarly here,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $ - \dfrac{1}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{1}{3} - 1}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{cosec x + cot x}}} \right)$

From the table of derivatives of trigonometric functions,

$

\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{cosec x) = - cosec(x)cot(x)}} \\

\dfrac{{\text{d}}}{{{\text{dx}}}}(\cot {\text{x) = - cose}}{{\text{c}}^2}({\text{x)}} \\

\\

$

Now,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $ - \dfrac{1}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{4}{3}}}\left( {{\text{ - cosec x cot x - cose}}{{\text{c}}^2}{\text{ x}}} \right)$

Take –cosec x common,

$\Rightarrow$ $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $\dfrac{{{\text{cosec x}}}}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{4}{3}}}\left( {{\text{cot x + cosec x}}} \right)$

Adding powers of similar terms, we get -------- (${{\text{a}}^{\text{m}}} \times {{\text{a}}^{\text{n}}} = {{\text{a}}^{{\text{m + n}}}}$)

$\Rightarrow$ $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $

\dfrac{{{\text{cosec x}}}}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{1}{3}}} \\

\\

$

= $\dfrac{{{\text{cosec x}}}}{{3{{\left( {{\text{cosec x + cot x}}} \right)}^{\dfrac{1}{3}}}}}$

Hence the answer.

Note: In order to solve these types of questions the key is to have a good idea on how to approach the differentiation of a wide variety of functions including trigonometric functions. Then with the help of the derivatives of cosec and cot functions the problem is further simplified. Then it’s all about rearranging the terms obtained using a few basic number properties to arrive at the answer.

Complete step-by-step answer:

Given Data,

y =$\dfrac{1}{{\sqrt[3]{{{\text{cosec x + cot x}}}}}}$

Transform y such that there is no cube root in the equation, for the ease of solving

$\Rightarrow$ y = $\dfrac{1}{{{{\left( {\cos {\text{ec x + cot x}}} \right)}^{\dfrac{1}{3}}}}} = {\left( {\cos {\text{ec x + cot x}}} \right)^{ - \dfrac{1}{3}}}$

Differentiating y with respect to x

$\Rightarrow$ $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{1}{3}}}$

For a function f = (x + 1)$^2$, $\dfrac{{{\text{df}}}}{{{\text{dx}}}}{\text{ becomes 2(x + 1}}{{\text{)}}^{2 - 1}}\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{x + 1)}}$

Similarly here,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $ - \dfrac{1}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{1}{3} - 1}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{cosec x + cot x}}} \right)$

From the table of derivatives of trigonometric functions,

$

\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{cosec x) = - cosec(x)cot(x)}} \\

\dfrac{{\text{d}}}{{{\text{dx}}}}(\cot {\text{x) = - cose}}{{\text{c}}^2}({\text{x)}} \\

\\

$

Now,

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $ - \dfrac{1}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{4}{3}}}\left( {{\text{ - cosec x cot x - cose}}{{\text{c}}^2}{\text{ x}}} \right)$

Take –cosec x common,

$\Rightarrow$ $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $\dfrac{{{\text{cosec x}}}}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{4}{3}}}\left( {{\text{cot x + cosec x}}} \right)$

Adding powers of similar terms, we get -------- (${{\text{a}}^{\text{m}}} \times {{\text{a}}^{\text{n}}} = {{\text{a}}^{{\text{m + n}}}}$)

$\Rightarrow$ $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ = $

\dfrac{{{\text{cosec x}}}}{3}{\left( {{\text{cosec x + cot x}}} \right)^{ - \dfrac{1}{3}}} \\

\\

$

= $\dfrac{{{\text{cosec x}}}}{{3{{\left( {{\text{cosec x + cot x}}} \right)}^{\dfrac{1}{3}}}}}$

Hence the answer.

Note: In order to solve these types of questions the key is to have a good idea on how to approach the differentiation of a wide variety of functions including trigonometric functions. Then with the help of the derivatives of cosec and cot functions the problem is further simplified. Then it’s all about rearranging the terms obtained using a few basic number properties to arrive at the answer.

Last updated date: 24th Sep 2023

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