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

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Answer 1 · 2016-11-27T01:47:01

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

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:
$-t dy/dt = 1/(2sqrt(t-1))$
$:. dy/dt = -1/(2sqrt(t-1)) * 1/t$
$:. dy/dt = -1/(2tsqrt(t-1))$

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