What is the derivative of #y=xarcsin(x^2)#?

To make the calculations more interesting, I'll assume that you don't know what the derivative of #arcsin(x)# is.

You can differentiate this function by using implicit differentiation. Start by isolating #arcsin(x^2)# on one side

#y/x = arcsin(x^2)" "color(blue)((1))#

This is equivalent to

#sin(y/x) = x^2" "color(blue)((2))#

Differentiate both sides with respect to #x#

#d/dx(sin(y/x)) = d/dx(x^2)#

#cos(y/x) * ((dy)/dx * 1/x - y/x^2) = 2x#

Rearrange this to get #(dy)/dx# isolated on one side

#(dy)/dx * cos(y/x) * 1/x - y/x^2 * cos(y/x) = 2x#

#(dy)/dx * cos(y/x) * 1/x = 2x + y/x^2 * cos(y/x)#

#(dy)/dx = (2x + y/x^2 * cos(y/x))/(cos(y/x) * 1/x)#

#(dy)/dx = (2x)/(cos(y/x) * 1/x) + y/x^color(red)(cancel(color(black)(2))) * color(red)(cancel(color(black)(x))) * color(red)(cancel(color(black)(cos(y/x))))/color(red)(cancel(color(black)(cos(y/x))))#

#(dy)/dx = (2x^2)/cos(y/x) * y/x#

Use the trigonometric identity

#color(blue)(sin^2x + cos^2x = 1)#

To write #cos(y/x)# as a function of #sin(y/x)#

#cos^2x = 1 - sin^2x#

#sqrt(cos^2x) = sqrt(1-sin^2x)#

#cosx = sqrt(1 - sin^2x)#

Use this identity and equations #color(blue)((1))# and #color(blue)((2))# to write

#(dy)/dx = (2x^2)/sqrt(1-sin^2(y/x)) + arcsin(x^2)#

#(dy)/dx = (2x^2)/sqrt(1 - (x^2)^2) + arcsin(x^2)#

#(dy)/dx = color(green)((2x^2)/sqrt(1-x^4) + arcsin(x^2))#