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What is the derivative of #pi^sinx#?

1 Answer

Jul 9, 2016

Letting #y=pi^(sin(x))#, we get

#y' = pi^(sin(x)) * cos(x) * ln(pi)# by the chain rule.

Explanation:

The derivatives of exponential functions with base #a# states that

#d/dx[a^(x)] = a^(x) * ln(a)#.

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