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  • 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)

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

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