Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 41 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 41 maths textbook solution

Answers (1)

Answer: \frac{d y}{d x}=\frac{\log (\operatorname{cosy})-y \cot x}{\log (\sin x)+x \tan y}

Hint:  To solve this equation we will convert log function

Given:  (\sin x)^{y}=(\cos y)^{x}

Solution:  

        \begin{aligned} &\log (\sin x)^{y}=\log (\cos y)^{x} \\\\ &y\log (\sin x)=x \log (\cos y) \end{aligned}

        \begin{aligned} &y \frac{d}{d x} \log (\sin x)+\log \sin x \frac{d y}{d x}=x \frac{d}{d x} \log \cos y+\log \cos y \frac{d x}{d x} \\\\ &y \frac{1}{\sin x} \cos x+\log \sin x \frac{d y}{d x}=x \frac{1}{\cos y}(-\sin x) \frac{d y}{d x}+\log \cos y \cdot 1 \end{aligned}

        \begin{aligned} &y \cot x+\log \sin x \frac{d y}{d x}=-x \tan \frac{d y}{d x}+\log \cos y \\\\ &\frac{d x}{d x} \log (\sin x)+y \frac{\cos x}{\sin x}=\log (\cos y)+x \frac{-\sin y}{\cos y} \end{aligned}

        \begin{aligned} &\frac{d x}{d x} \log (\sin x)+x \tan y=\log (\cos y)-y \cot x \\\\ &\frac{d y}{d x}=\frac{\log (\cos y)-y \cot x}{\log (\sin x)+x \tan y} \end{aligned}

Hence it is proved

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16
anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8
anam-khan

CBSE private candidates challenge cancellation of post-Class 12 additional subject exam in Delhi HC

Latest Question