Question

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

Answer 1

`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))`