Q6    Differentiate w.r.t. x the function in Exercises 1 to 11.

         \cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2

Answers (1)

Given function is
f(x)=\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2
Now, rationalize the [] part
\left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ]= \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} .\frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}\right ]
                                                       
                                                          =\frac{(\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x })^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2} \ \ \ \ \ \ (Using \ (a-b)(a+b)=a^2-b^2)
                                                       
                                                          =\frac{((\sqrt { 1+ \sin x })^2+ (\sqrt { 1- \sin x })^2+2(\sqrt { 1+ \sin x })(\sqrt { 1- \sin x }))}{1+\sin x-1+\sin x}
                                                                                                                                                                         (Using \ (a+b)^2=a^2+b^2+2ab)
                                                          =\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x} }{2\sin x}
                                                         
                                                           =\frac{2(1+\cos x)}{2\sin x} = \frac{1+\cos x}{\sin x}

                                                         =\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \ \ \ \ \ (\because 2\cos^2= 1+\cos2x \ and \ \sin2x = 2\sin x\cos x)
                                                   
                                                         =\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot \frac{x}{2}
Given function reduces to
f(x) = \cot^{-1}(\cot \frac{x}{2})\\ f(x) = \frac{x}{2}
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{x}{2})}{dx} = \frac{1}{2}
Therefore, differentiation w.r.t. x is   \frac{1}{2}

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions