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 25 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 25 maths textbook solution

Answers (1)

best_answer

Answer: \frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]+(\sin x)^{x}(\log \sin x)+x \cdot \cot x

Hint: Differentiate the equation taking log on both sides

Given: y=x^{\sin x}+(\sin x)^{x}

Solution:  

\begin{aligned} &\text { Let } u=x^{\sin x} \text { and } v=(\sin x)^{x} \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\\\ &u=x^{\sin x} \end{aligned}

\begin{aligned} &\log u=\log x^{\sin x} \\\\ &\frac{d}{d x}(\log u)=\frac{d}{d x}(\sin x \cdot \log x) \end{aligned}

\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \frac{d}{d x}(\log x) \quad\left[\because \frac{d}{d x}(\log x)=\frac{1}{x}\right]

\frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right] \quad\left[\because \frac{d}{d x}(\sin x)=\cos x\right]

\begin{aligned} &v=(\sin x)^{x} \\\\ &\log v=\log (\sin x)^{x} \\\\ &\frac{d}{d x} v=\frac{d}{d x}[x \log (\sin x)] \end{aligned}

\frac{1}{v} \frac{d v}{d x}=\cdot \log (\sin x)+x \cdot \frac{d}{d x}(\log \sin x)

          =\log (\sin x)+x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)

\begin{aligned} &\frac{d v}{d x}=(\sin x)^{x}[\log \sin x+x \cdot \cot x] \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \end{aligned}

\frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]+(\sin x)^{x}(\log \sin x)+x \cdot \cot x

Posted by

infoexpert26

View full answer

Similar Questions

Latest Question