Question

How do you find the derivative of `g(alpha)=5^(-alpha/2)sin2alpha`?

Answer 1

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

Explanation:

The derivative of `b^u`, 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..