How do you find the derivative of the function: #y=arccos(1/x)#?

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Answer 1 · 2016-11-08T21:44:03

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

We can write #y=arccos(1/x) <=>cosy=1/x#
# :. cosy=x^-1 #

We can then differentiate implicitly:

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

Using the identity # sin^2A + cos^2A -= 1 # we have

# sin^2y + (1/x)^2 = 1 #
# :. sin^2y = 1 - 1/x^2 #
# siny = sqrt(1 - 1/x^2) #

And substituting this we have:

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