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Differentiation — Maths STD 12 Science — Question

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

Answer

Let $\text{y}=(\sin^{-1}\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin^{-1}\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(\sin^{-1}\text{x})\ \big[\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\sin^{-1}\text{x})+\log\sin(-1)\times\frac{\text{d}}{\text{dx}}(\text{x})$
$=\text{x}\times\frac{1}{\sin^{-1}\text{x}}\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)+\log\sin^{-1}\text{x}(1)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\sin^{-1}\text{x}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\log\sin^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\bigg[\log\sin^{-1}\text{x}+\frac{\text{x}}{\sin^{-1}\text{x}\big(\sqrt{1-\text{x}^2}\big)}\bigg]$
$\frac{\text{dy}}{\text{dx}}=(\sin^{-1}\text{x})^2\Big[\log\sin^{-1}\text{x}+\frac{\text{x}}{\sin^{-1}\text{x}\sqrt{1-\text{x}^2}}\Big]$
[Using equation (i)]

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