Differentiate ( 2x)^ x + ^ -1 3x with respect to x. - Vidyadip

Question

Differentiate $(\sin 2x)^{x} + \sin^{-1}\sqrt{3x}$ with respect to $x.$

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Answer

$\text{y} = (\sin 2{x})^{x} + \sin^{-1}(\sqrt{3x)} = u + v$
$\therefore\frac{dy}{dx} = \frac{du}{dx}+ \frac{dv}{dx}$
$u = (\sin 2x)^{x}\Rightarrow \log u = x\log \sin^{2}x$
$\frac{1}{u}\frac{du}{dx} = 2x. \cot 2x + \log \sin 2x$
$\therefore\frac{du}{dx} = (\sin 2x)^{x}[2x \cot 2x + \log \sin 2x]$
$\frac{dv}{dx} = \frac{1}{\sqrt{1 - 3x}}\frac{\sqrt{3}}{2\sqrt{x}}$
$\therefore\frac{dy}{dx} = (\sin 2x)^x[2x \cot 2x + \log \sin 2x] +\frac{\sqrt{3}}{2\sqrt{x} \sqrt{1 - 3x}}$

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