What is the derivative of #Y = Cos^-1 (e^-T)#?
1 Answer
Mar 22, 2018
# (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)) #
