How do you differentiate #y=sin^-1sqrtx#?

1 Answer
Steve M
Nov 23, 2016

# dy/dx = 1/(2sqrtxsqrt(1 -x)) #

Explanation:

# y= sin^-1sqrtx <=> siny=x^(1/2) #

Differentiating (Implicitly) wrt #x#:

# cosydy/dx = 1/2x^(-1/2) #
# :. cosydy/dx = 1/(2sqrtx) #

Using the Identity #sin^y+cos^2y -= 1#
# :. (x^(1/2))^2 + cos^2y = 1 #
# :. x + cos^2y = 1 #
# :. cos^2y = 1 -x#
# :. cosy = sqrt(1 -x) #

And so:
# :. sqrt(1 -x)dy/dx = 1/(2sqrtx) #
# :. dy/dx = 1/(2sqrtxsqrt(1 -x)) #

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