How do you find the derivative of g(alpha)=5^(-alpha/2)sin2alphag(α)=5α2sin2α?

1 Answer
Feb 1, 2018

[5^(-a/2) * ln(5)* -1/2sin(2a)] + [5^(-a/2)*2cos(2x)][5a2ln(5)12sin(2a)]+[5a22cos(2x)]
You could probably simplify this more though

Explanation:

The derivative of b^ubu, where b is a number and u is basically anything is: b^u * ln(b) * u', which I memorized with something like "bowling boy"
So we have to use product rule for this, so we need the derivatives of both 5^(-a/2 and sin(2a). Using the formula above, the derivative of the first is 5^(-a/2) * ln(5)* -1/2, and use chain rule to derive the second to get 2cos(2x). Plug these into the product rule and you get the answer, which is probably able to be simplified further..