
How do you find the derivative of ${\csc ^{ - 1}}(u)$?
Answer
531.6k+ views
Hint: The derivative of the inverse function given here can be found by differentiating after transforming the function to a convenient variable that will then be easily differentiable. For the achieving the above given task we first change the ${\csc ^{ - 1}}(u)$ into $\csc x$ and then we will differentiate it because we know the standard formula for it which is $\dfrac{d}{{dx}}(\csc x) = - \csc x\cot x$
The function then will be substituted back for the value of $u$and then we will find the answer in the terms of the given variable which here is given as $u$. For the above question we should also know one basic trigonometric identity which is the relation between the $\csc x$and $\cot x$that particular identity goes as follows:
${\csc ^2}x - {\cot ^2}x = 1$ which is a very standard identity that relates the above two functions.
Complete step-by-step answer:
First we convert our given function into a function of $x$and after solving we will substitute it back to the original variable which is $u$. For that let,
$x = {\csc ^{ - 1}}u$
Which will then give
$u = \csc x$
the above function will now be differentiated by $x$as follows:
\[\dfrac{{du}}{{dx}} = - \csc x\cot x\]
We will now substitute back the variable, first we will write
\[cotx = \sqrt {{u^2} - 1} \]
then we will substitute all the values of $x$ back to $u$
we get,
\[\dfrac{{du}}{{dx}} = - u\sqrt {{u^2} - 1} \]
But since we had to find \[\dfrac{{dx}}{{du}}\]
we will write,
\[\dfrac{{dx}}{{du}} = - \dfrac{1}{{u\sqrt {{u^2} - 1} }}\]
Which will be our final answer.
So, the correct answer is “\[- \dfrac{1}{{u\sqrt {{u^2} - 1} }}\]”.
Note: In calculating such types of questions the equation should always be converted into the variable it will be easily solvable in by using the standard formula and then transforming back to its original variable. Also remember the basic trigonometric identities.
The function then will be substituted back for the value of $u$and then we will find the answer in the terms of the given variable which here is given as $u$. For the above question we should also know one basic trigonometric identity which is the relation between the $\csc x$and $\cot x$that particular identity goes as follows:
${\csc ^2}x - {\cot ^2}x = 1$ which is a very standard identity that relates the above two functions.
Complete step-by-step answer:
First we convert our given function into a function of $x$and after solving we will substitute it back to the original variable which is $u$. For that let,
$x = {\csc ^{ - 1}}u$
Which will then give
$u = \csc x$
the above function will now be differentiated by $x$as follows:
\[\dfrac{{du}}{{dx}} = - \csc x\cot x\]
We will now substitute back the variable, first we will write
\[cotx = \sqrt {{u^2} - 1} \]
then we will substitute all the values of $x$ back to $u$
we get,
\[\dfrac{{du}}{{dx}} = - u\sqrt {{u^2} - 1} \]
But since we had to find \[\dfrac{{dx}}{{du}}\]
we will write,
\[\dfrac{{dx}}{{du}} = - \dfrac{1}{{u\sqrt {{u^2} - 1} }}\]
Which will be our final answer.
So, the correct answer is “\[- \dfrac{1}{{u\sqrt {{u^2} - 1} }}\]”.
Note: In calculating such types of questions the equation should always be converted into the variable it will be easily solvable in by using the standard formula and then transforming back to its original variable. Also remember the basic trigonometric identities.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

