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 29 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 29 sub question (i) maths textbook solution

Answers (1)

best_answer

Answer: \frac{d y}{d x}=x \cos x\left(\frac{\cos x}{x}-\sin x \log x\right)+\sin ^{\tan x}\left(1+\sec ^{2} x \log \sin x\right)

Hint: To solve this equation we use f(x)^{f(x)}  formula

Given:

Solution:  

\begin{gathered} y=x^{\cos x}+\sin x^{\tan x} \\\\ y=(f(x))^{f(x)} \end{gathered}

\begin{aligned} &\text { Let } u=x^{\cos x} \\\\ &\log u=x^{\cos x} \\\\ &\log u=\cos x \log x \end{aligned}

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

\begin{aligned} &=x \cos x\left[\left(\frac{\cos x}{x}-\sin x \log x\right)\right] \\\\ &v=\sin x^{\tan x} \\\\ &\log v=\tan x \log \sin x \end{aligned}

\frac{1}{v} \frac{d v}{d x}=\tan x\left[\frac{1}{\sin x} \frac{d}{d x}(\sin x)\right]+\log \sin x \sec ^{2} x

\frac{d v}{d x}=v\left(\tan x \times \frac{\cos x}{\sin x}+\sec ^{2} x \log \sin x\right)

\frac{d v}{d x}=v\left(\frac{\sin x}{\cos x} \times \frac{\cos x}{\sin x}+\sec \sin ^{2} x \log \sin x\right)

\frac{d v}{d x}=\sin x^{\tan x}\left(1+\sec ^{2} x \log \sin x\right)

\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}

\frac{d v}{d x}=x \cos x\left[\left(\frac{\cos x}{x}-\sin x \log x\right)\right]+\sin x^{\tan x}\left ( 1+\sec ^{2} x\log \sin x\right )

 

 

 

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 Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question