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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:
$\text{x}^{\cos^{-1}\text{x}}$
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Answer

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

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