What is the derivative of this function #y = sin ^ 1 (x^2)?
1 Answer
Aug 13, 2017
Explanation:
We're asked to find the derivative
#(dy)/(dx) [sin^1(x^2)]#
Let's use the chain rule first:
#d/(dx) [sin^1(x^2)] = d/(du) [sin^1u] (du)/(dx)#
where

#u = x^2# 
#d/(du) [sin^1u] = 1/(sqrt(1u^2))# :
#y'(x) = (d/(dx)[x^2])/(sqrt(1x^4))#
The derivative of
#color(blue)(ulbar(stackrel(" ")(" "y'(x) = (2x)/(sqrt(1x^4))" "))#