Question

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

Answer 1

`y=tan(arcsin(x))=(sin(arcsin(x))/(cos(arcsin(x)))\\\\We\ know:\\sin(arcsin(x))=x\ and\ cos(arcsin(x))=√(1-x^2)\\\\therefore:y=(x)/(√(1-x^2))\\\\y'=\left((x)/(√(1-x^2))\right)'=(x'√(1-x^2)-x(√(1-x^2))')/((√(1-x^2))^2)=(√(1-x^2)-x\cdot(1)/(2√(1-x^2))\cdot(-2x))/(1-x^2)\\\\=(√(1-x^2)+(x^2)/(√(1-x^2)))/(1-x^2)=((1-x^2+x^2)/(√(1-x^2)))/(1-x^2)=(1)/((1-x^2)√(1-x^2))=(1)/((1-x^2)(1-x^2)^(1)/(2))`

`.\center\boxed{=(1)/((1-x^2)^(3)/(2))}`

Answer 2

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