Question

How do you find the derivative of y=tan(arcsin(x))y=tan(arcsin(x)) ?

Answer 1

Answer is in the attachment below. If you have any questions about the workings, just leave a comment below.

Answer 2

$y=tan(arcsin(x))=\frac{sin(arcsin(x)}{cos(arcsin(x))}\\\\We\ know:\\sin(arcsin(x))=x\ and\ cos(arcsin(x))=\sqrt{1-x^2}\\\\therefore:y=\frac{x}{\sqrt{1-x^2}}\\\\y'=\left(\frac{x}{\sqrt{1-x^2}}\right)'=\frac{x'\sqrt{1-x^2}-x(\sqrt{1-x^2})'}{(\sqrt{1-x^2})^2}=\frac{\sqrt{1-x^2}-x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot(-2x)}{1-x^2}\\\\=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}=\frac{\frac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}=\frac{1}{(1-x^2)\sqrt{1-x^2}}=\frac{1}{(1-x^2)(1-x^2)^\frac{1}{2}}$
$.\center\boxed{=\frac{1}{(1-x^2)^\frac{3}{2}}}$