What is the derivative of # ArcSin[x^(1/2)]#?

1 Answer
Dec 4, 2016

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

Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you use the chain rule.

Let #y=arcsin1(x^(1/2)) <=> siny=sqrt(x) #

Differentiate Implicitly:

# cosydy/dx = 1/2x^(-1/2) #
# :. cosydy/dx = 1/(2sqrt(x)) # ..... [1]

Using the #sin"/"cos# identity;

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

Substituting into [1]
# :. sqrt(1-x)dy/dx = 1/(2sqrt(x)) #
# :. dy/dx = 1/(2sqrt(x)sqrt(1-x)) #