How do you find the derivative of #arcsin(x^2/4)#?

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Answer 1 · 2016-11-15T19:06:58

# d/dxarcsin(x^2/4) = (x)/(2sqrt(1-x^2/16)) #

# y = arcsin(x^2/4) #

# siny = x^2/4 # ..... [1]

We can now differentiate implicitly to get:

# cos(y)dy/dx = x/2 # ..... [2]

Using the fundamental trig identity #sin^2A+cos^2A-=1# we can write:

# sin^2(y)+cos^2(y)=1#
# :. (x^2/4)^2+cos^2(y)=1# (from [1])
# :. cos^2(y)=1-x^2/16#
# :. cos(y)=sqrt(1-x^2/16)#

Substituting into [2] we get:

# sqrt(1-x^2/16)dy/dx = x/2 #

# :. dy/dx = (x)/(2sqrt(1-x^2/16)) #