How do you differentiate #y=cos^-1(1-3x^2)#?

1 Answer
Steve M
Oct 26, 2016

# dy/dx=(6x)/sqrt(6x-9x^2) #

Explanation:

Easiest method is to rewrite;

# y=cos^-1(1-3x^2) # as # cosy=1-3x^2 #

and to use implicit differentiation to give:
#-sinydy/dx =-6x #
# :. sinydt/dx =6x #
# :. dy/dx =(6x)/siny #

And using # cos^2A+sin^2A-=1 => siny =sqrt(1-cos^2y) #
# :. siny =sqrt(1-(1-3x^2)^2) #
# :. siny =sqrt(1-(1-3x^2)^2) #
# :. siny =sqrt(1-(1-6x+9x^2)) #
# :. siny =sqrt(1-1+6x-9x^2) #
# :. siny =sqrt(6x-9x^2) #

And so, # dy/dx=(6x)/sqrt(6x-9x^2) #

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