CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x: $\text{x}^{\tan^{-1}\text{x}}$
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Answer
Let $\text{y}=\text{x}^{\tan^{-1}\text{x}}\ .....(\text{i})$ Taking log on both the sides, $\log\text{y}=\log\text{x}^{\tan^{-1}\text{x}}$ $\log\text{y}=\tan^{-1}\text{x}\log\text{x}\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$ Differentiating it with respect to x using product rule, $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\tan^{-1}\text{x})$ $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}\Big(\frac{1}{1+\text{x}^2}\Big)$ $\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$ $\frac{\text{dy}}{\text{dx}}=\text{x}^{\tan^{-1}\text{x}}\Big[\frac{\tan^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{1+\text{x}^2}\Big]$ [Using equation (i)]
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