I am having a hard time in Calculus 1. Please help :)

I need to obtain the derivative of:
p(θ) = (π/2)^ Sinθ
Please be clear with your answer, and how you got to it so I may understand :)

It's easier if you take log of both sides, then differentiate:

p(θ) = (π/2)^sinθ
ln(p(θ)) = sinθ ln(π/2)
1/p(θ) * p'(θ) = ln(π/2) cosθ

p'(θ) = p(θ) ln(π/2) cosθ
p'(θ) = (π/2)^sinθ * ln(π/2) cosθ

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General rule:

d/dx (c^f(x)) = c^f(x) * ln(c) f'(x), where c is some constant

derivative is : (π/2)^ Sinθ * ln( (π/2) ) * Cosθ

learn these: http://en.wikipedia.org/wiki/Differentia… ... and then apply them (Chain rule)
and: http://en.wikipedia.org/wiki/Exponential…

The formula you need is d(a^u)= (a^u)(ln a)(du)
This is used when a constant is raised to a function

a=pi/2, u = sin (theta), du = cos (theta)
So p'(theta ) =[ (pi/2)^(sin(theta))](ln(pi/2)(cos(theta)