Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\text{x}^{\tan^{-1}\text{x}}$
✓
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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