Question
Differentiate the following functions with respect to x:
$\text{x}^{\sin^{-1}\text{x}}$
$\text{x}^{\sin^{-1}\text{x}}$
✓
Answer
Let $\text{y}=\text{x}^{\sin^{-1}\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin^{-1}\text{x}}$
$\log\text{y}=\sin^{-1}\text{x}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+(\log\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin^{-1}\text{x}}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
Taking log on both the sides,
$\log\text{y}=\log\text{x}^{\sin^{-1}\text{x}}$
$\log\text{y}=\sin^{-1}\text{x}\log\text{x}$
Differentiating it with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+(\log\text{x})\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}\Big(\frac{1}{\text{x}}\Big)+(\log\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin^{-1}\text{x}}\Big[\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{\log\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]
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