Question

$y=\frac{e^{3x}(si{n}^{-1}x{)}^4}{\sqrt{1-x^2}si{n}^{3}x}$ find $\frac{dy}{dx}$

Answer 1

We have,

$y=\frac{e^{3x}{\left({\sin }^{-1}x\right)}^4}{\sqrt{1-x^2}{\sin }^{3}x}$

On differentiating and we get,

$\frac{dy}{dx}=\frac{\sqrt{1-x^2}{\sin }^{3}x\frac{d}{dx}{e}^{3x}{\left({\sin }^{-1}x\right)}^4-e^{3x}{\left({\sin }^{-1}x\right)}^4\frac{d}{dx}\sqrt{1-x^2}{\sin }^{3}x}{{\left[\sqrt{1-x^2}{\sin }^{3}x\right]}^2}$

$=\frac{\sqrt{1-x^2}{\sin }^{3}x[e^{3x}\frac{d}{dx}{\left({\sin }^{-1}x\right)}^4+{\left({\sin }^{-1}x\right)}^4\frac{d}{dx}{e}^{3x}]-e^{3x}{\left({\sin }^{-1}x\right)}^4[\sqrt{1-x^2}\frac{d}{dx}{\sin }^{3}x+{\sin }^{3}x\frac{d}{dx}\sqrt{1-x^2}]}{{\left[\sqrt{1-x^2}{\sin }^{3}x\right]}^2}$

$=\frac{\sqrt{1-x^2}{\sin }^{3}x[e^{3x}4{\left({\sin }^{-1}x\right)}^3\frac{1}{\sqrt{1-x^2}}+{\left({\sin }^{-1}x\right)}^{4}3e^{3x}]-e^{3x}{\left({\sin }^{-1}x\right)}^4[\sqrt{1-x^2}3{\sin }^{2}x\mathrm{cosx}+{\sin }^{3}x\frac{1}{2\sqrt{1-x^2}}(-2x)]}{{\left[\sqrt{1-x^2}{\sin }^{3}x\right]}^2}$

hence, this is the answer.