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
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) #