Answer
Verified
478.2k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE