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

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

Answers (1)

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

Hint: \text { Diff by }\left(x+\frac{1}{x}\right)^{x}

Given: \left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}

Solution:   \text { Let } y=\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}

        y=u+v

Diff w.r.t x

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

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

Calculate \frac{d u}{d x}, u=\left(x+\frac{1}{x}\right)^{x}

Take log on both side

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

        \frac{1}{u} \frac{d u}{d x}=\frac{d\left(x \log \left(1+\frac{1}{x}\right)\right.}{d x}

Use product rule (u v)^{\prime}=u^{\prime} v+v^{\prime} u

        \frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\frac{1}{\left(x+\frac{1}{x}\right)} \frac{d}{d x}\left(x+\frac{1}{x}\right) x

        \frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\left(\frac{x}{x^{2}+1}\left(\frac{x^{2}-1}{x^{2}}\right) x\right)

        \frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\left(\frac{x^{2}-1}{x^{2}+1}\right)

        \frac{d u}{d x}=u\left(\log \left(x+\frac{1}{x}\right)\right)+\left(\frac{x^{2}-1}{x^{2}+1}\right)

        \frac{d u}{d x}=\left(x+\frac{1}{x}\right)^{x} \cdot\left(\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right)

Calculate  \frac{d v}{d x}

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

Taking \log v=\left[1+\frac{1}{x}\right] \log x

Diff w.r.t. x

Use product rule

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

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

        \frac{1}{v}\left(\frac{d v}{d x}\right)=\frac{-1}{x^{2}} \cdot \log x+\frac{1}{x}\left(1+\frac{1}{x}\right)

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

                                                                    \frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)

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

Put value of  \frac{d u}{d x} \text { and } \frac{d v}{d x}

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

Posted by

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