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Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 20 maths

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Answer:  \frac{d y}{d x}=n \cdot x^{n-1}+n^{x} \cdot \log n+x^{x}(\log x+1)

Hint:   Differentiate the equation taking log on both sides

Given: y=x^{x}+n^{x}+x^{x}+n^{n}

Solution:  y=x^{x}+n^{x}+x^{x}+n^{n}

As we know

        \left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]

        \left[\frac{d}{d x}(\log x)=\frac{1}{x}\right]

According to question

        \frac{d y}{d x}=\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(n^{x}\right)+\frac{d u}{d x}+\frac{d}{d x}\left(x^{n}\right)        .......(1)

Take u=x^{x}

        \begin{aligned} \frac{d}{d x}(\log u) &=\log x^{x} \\ &=\frac{d}{d x}(x \log x) \end{aligned}

         \begin{aligned} &\frac{1}{u} \frac{d u}{d x}=\frac{d x}{d x} \cdot \log x+x \frac{d}{d x}(\log x) \\\\ &\frac{d u}{d x}=u\left[\log +\not{x} \cdot \frac{1}{\not{x}}\right] \\\\ &\frac{d u}{d x}=x^{x}[\log x+1] \end{aligned}\

Put in eq. (1)

        \frac{d y}{d x}=n \cdot x^{n-1}+n^{x} \cdot \log n+x^{x}(\log x+1)


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