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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 3

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 3

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Answer: (1+\cos x)^{x}\left[\left(-\frac{x \sin x}{1+\cos x}\right)+\log (1+\cos x)\right.

Hint: Differentiate by \cos ^{n}x

Given: (1+\cos x)^{x}

Solution:Let y=(1+\cos x)^{x}            ...........(i)

Taking log on both the sides,

        \begin{aligned} &\log y=\log (1+\cos x)^{x} \\\\ &\log y=x \log (1+\cos x) \end{aligned}

Differentiating with respect to x,

        \frac{1}{y} \frac{d y}{d x}=x \frac{d}{d x} \log (1+\cos x)+\log (1+\cos x) \frac{d}{d x}(x)        [Using product rule]

        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=x \frac{1}{(1+\cos x)} \frac{d}{d x}(1+\cos x)+\log (1+\cos x)(1) \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{x}{(1+\cos x)}(0-\sin x)+\log (1+\cos x) \end{aligned}

        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\log (1+\cos x)-\frac{x \operatorname{som} x}{(1+\cos x)} \\\\ &\frac{d y}{d x}=y\left[\log (1+\cos x)-\frac{x \sin x}{1+\cos x}\right] \end{aligned}

        \frac{d y}{d x}=(1+\cos x)^{x}\left[\log (1+\cos x)-\frac{x \sin x}{(1+\cos x)}\right]        [Using equation (i)]

 

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