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  • Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 18 sub question (vii) maths

Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 18 sub question (vii) maths

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

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

Solution:  y=(\cos x)^{x}+(\sin x)^{\frac{1}{x}}
                    =u+v

                \begin{aligned} &u=(\cos x)^{x} \text { and } \\\\ &v=(\sin x)^{\frac{1}{x}} \end{aligned}

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

                \frac{d u}{d x}=(\cos x)^{x}(\ln \cos x-x \tan x)        ...........(1)

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

                \begin{aligned} \frac{d v}{d x} &=\frac{(\sin x)^{\frac{1}{x}}}{x^{2}}[x \cot x-\log (\sin x)] \\\\ y &=u+v \end{aligned}

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

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

 

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