Question

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

Answer 1

`y^' = (2x^2)/sqrt(1-x^4) + arcsin(x^2)`

Explanation:

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))`