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

##### 1 Answer

# 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 "

# 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

Using

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

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

**NOTE - Derivative of #arcsin(x)#**

We can easily find

Let

# :. 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) #