How do you differentiate #y=cot^-1sqrt(t-1)#?

1 Answer
Steve M
Nov 27, 2016

# dy/dt = -1/(2tsqrt(t-1)) #

Explanation:

Let # y = cot^-1sqrt(t-1) => coty=sqrt(t-1)#

Differentiating implicitly, and applying the chain rule:
# -csc^2y dy/dt = 1/2(t-1)^(-1/2) #
# -csc^2y dy/dt = 1/(2sqrt(t-1) #

Using the identity #1 + cot^2A -= csc^2A # we can write;
# 1+cot^2y=csc^2y #
# :. 1+(sqrt(t-1))^2=csc^2y #
# :. csc^2y=1+(t-1) #
# :. csc^2y=t #

And so:
# -t dy/dt = 1/(2sqrt(t-1)) #
# :. dy/dt = -1/(2sqrt(t-1)) * 1/t#
# :. dy/dt = -1/(2tsqrt(t-1)) #

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
1697 views around the world
You can reuse this answer
Creative Commons License