Question

$\large \bf \orange {Namaste} \: \red{ Everyone }$
Find dy/dx
If
$\: \: \: \: \: \: \: \: \boxed{ \sf y = x \times {sin}^{ - 1} x}$
​

Answer 1

$\tt y = x{sin}^{-1}x \\ \\ \tt \boxed{\bold{\underline{\green{\tt Multiplication\:Rule: \frac{d(uv)}{dx} = \frac{du}{dx}.v + \frac{dv}{dx}.u }}}} \\ \\ \tt \therefore \frac{dy}{dx} = \frac{dx}{dx}.{sin}^{-1}x + \frac{d({sin}^{-1}x)}{dx}.x \\ \\ \therefore \tt \frac{dy}{dx} = {sin}^{-1}x + \frac{x}{\sqrt{1+{x}^{2}}} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt \frac{dy}{dx} = {sin}^{-1}x + \frac{x}{\sqrt{1+{x}^{2}}} }}}}$

Answer 2

Topic :-

Differentiation

To Differentiate :-

$\sf{y=x\sin^{-1}x}$

Solution :-

$\sf{y=x\sin^{-1}x}$

$\sf{\dfrac{dy}{dx}=\dfrac{d(x\sin^{-1}x)}{dx}}$

$\sf{\dfrac{dy}{dx}=\sin^{-1}x\cdot\dfrac{dx}{dx}+x\cdot\dfrac{d(\sin^{-1}x)}{dx}}$

$\sf {\left(\because \dfrac{d(fg)}{dx}= g\dfrac{df}{dx}+f\dfrac{dg}{dx} \right)}$

$\sf{\dfrac{dy}{dx}=\sin^{-1}x+x\cdot\dfrac{d(\sin^{-1}x)}{dx}}$

$\sf {\left(\because \dfrac{dx}{dx}=1\right)}$

$\sf{\dfrac{dy}{dx}=\sin^{-1}x+x\cdot\dfrac{1}{\sqrt{1-x^2}}}$

$\sf {\left(\because \dfrac{d(\sin^{-1}x)}{dx}=\dfrac{1}{\sqrt{1-x^2}}\right)}$

$\sf{\dfrac{dy}{dx}=\sin^{-1}x+\dfrac{x}{\sqrt{1-x^2}}}$

Answer :-

$\boxed{\sf{\dfrac{dy}{dx}=\sin^{-1}x+\dfrac{x}{\sqrt{1-x^2}}}}$

Additional Formulae :-

$\sf{\dfrac{d(\cos^{-1}x)}{dx}=\dfrac{-1}{\sqrt{1-x^2}}}$

$\sf{\dfrac{d(\tan^{-1}x)}{dx}=\dfrac{1}{1+x^2}}$

$\sf{\dfrac{d(\sec^{-1}x)}{dx}=\dfrac{1}{|x|\sqrt{x^2-1}}}$

$\sf{\dfrac{d(\csc^{-1}x)}{dx}=\dfrac{-1}{|x|\sqrt{x^2-1}}}$

$\sf{\dfrac{d(\cot^{-1}x)}{dx}=\dfrac{-1}{1+x^2}}$