Question

How do you differentiate `(arcsinx)^2`?

Answer 1

`d/dx (arcsinx)^2 = 2arcsinx/sqrt(1-x^2)`

Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If `y=f(x)` then `f'(x)=dy/dx=dy/(du)(du)/dx`

I was taught to remember that the differential can be treated like a fraction and that the "`dx`'s" of a common variable will "cancel" (It is important to realise that `dy/dx` isn't a fraction but an operator that acts on a function, there is no such thing as "`dx`" or "`dy`" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

`dy/dx = dy/(dv)(dv)/(du)(du)/dx` etc, or `(dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx)`

So with `y = (arcsinx)^2`, Then:

`{ ("Let "u=arcsinx, => , (du)/dx=1/sqrt(1-x^2) "(see below)"), ("Then "y=u^2, =>, dy/(du)=2u ) :}`

Using `dy/dx=(dy/(du))((du)/dx)` we get:

`dy/dx = (2u)(1/sqrt(1-x^2))`
`:. dy/dx = 2arcsinx/sqrt(1-x^2)`

NOTE - Derivative of `arcsin(x)`
We can easily find `d/dx(arcsinx)` using implicit differentiation:
Let `y = arcsinx <=> siny=x`

`:. cosydy/dx=1`
Using `sin^2y+cos^2y-=1 => cos^2y=1-x^2`
`:. cosy=sqrt(1-x^2)`
`:. sqrt(1-x^2)dy/dx=1`
`:. dy/dx=1/sqrt(1-x^2)`