Question

What is the derivative of `Y = Cos^-1 (e^-T)`?

Answer 1

`(dY)/(dT) = (e^(-T))/sqrt(1-e^(-2T))`

Explanation:

We use the following derivatives:

`{: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (af(x), af'(x),a " constant"), (e^(ax), ae^(ax), a " constant)"), (cos^(-1)x, -1/sqrt(1-x^2), ), (f(g(x)), f'(g(x)) \ g'(x),"(Chain rule)" ) :}`

So that:

`(dY)/(dT) = d/(dT) cos^(-1)(e^(-T))`

`\ \ \ \ \ \ = d/(dT) arccos(e^(-T))`

`\ \ \ \ \ \ = -1/sqrt(1-(e^(-T))^2) * d/(dT)( e^(-T))`

`\ \ \ \ \ \ = -(-e^(-T))/sqrt(1-(e^(-T))^2)`

`\ \ \ \ \ \ = (e^(-T))/sqrt(1-e^(-2T))`