Question

if y=sin^-1(3x/2-x^3/2) find dy/dx
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Answer 1

Answer:

if y=sin‹-1 then (3x/2-x/2) then you should find dy/dx

Answer 2

given,

$y=sin^-1(3x/2-x^3/2)\\=sin^{-1}\left(\frac{3x}{\frac{2-x^3}{2}}\right)\\=\frac{d}{dx}\left(\arcsin \left(\frac{6x}{2-x^3}\right)\right)\\apply the chain rule \\=\frac{1}{\sqrt{1-\left(\frac{6x}{2-x^3}\right)^2}}\frac{d}{dx}\left(\frac{6x}{2-x^3}\right)\\=\frac{1}{\sqrt{1-\left(\frac{6x}{2-x^3}\right)^2}}\cdot \frac{6\left(2+2x^3\right)}{\left(2-x^3\right)^2}\\=\frac{6\sqrt{\left(2-x^3\right)^2}\left(2+2x^3\right)}{\left(2-x^3\right)^2\sqrt{\left(2-x^3\right)^2-36x^2}}$