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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 18 sub question (ii)

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 18 sub question (ii)

Answers (1)

Answer: x^{\sin x-\cos x}\left[\frac{\sin x-\cos x}{x}+\log x(\sin x+\cos x)\right]+\frac{4 x}{\left(x^{2}+1\right)^{2}}

Hint: \text { Diff by } x^{n}

Given: y=x^{\sin x-\cos x}+\frac{x^{2}-1}{x^{2}+1}

Solution:  

          y=u+v

        \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}                .........(1)

u=x^{\sin x-\cos x}

Take log

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

        \begin{aligned} &\frac{d u}{d x}=\left[\frac{\sin x-\cos x}{x}+\log x(\sin x+\cos x) x^{\sin x-\cos x}\right. \\\\ &v=\frac{x^{2}-1}{x^{2}+1} \end{aligned}

        \begin{aligned} &\log v=\log \left(\frac{x^{2}-1}{x^{2}+1}\right) \\\\ &=\log \left(x^{2}-1\right)-\log \left(x^{2}+1\right) \end{aligned}

        \begin{aligned} &\frac{1}{v} \frac{d v}{d x}=\frac{1}{x^{2}-1}(2 x) \cdot \frac{-1}{x^{2}+1}(2 x) \\\\ &\frac{d v}{d x}=\left(\frac{2 x}{x^{2}+1}\right)\left(\frac{-2 x}{x^{2}+1}\right)\left(\frac{x^{2}-1}{x^{2}+1}\right) \end{aligned}

              \begin{aligned} &=2 \times\left(\frac{\left(x^{2}+1\right)-\left(x^{2}-1\right)}{\left(x^{2}-1\right)\left(x^{2}+1\right)}\right) \\\\ &=\frac{2(2 x)}{\left(x^{2}+1\right)^{2}} \end{aligned}

              =\frac{4 x}{\left(x^{2}+1\right)^{2}}

Using (1) we get

\frac{d y}{d x}=x^{\sin x-\cos x}\left[\frac{\sin x-\cos x}{x}+\log x(\sin x+\cos x)\right]+\frac{4 x}{\left(x^{2}+1\right)^{2}}

 

 

Posted by

infoexpert26

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