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

$\left[{5}^{- \frac{a}{2}} \cdot \ln \left(5\right) \cdot - \frac{1}{2} \sin \left(2 a\right)\right] + \left[{5}^{- \frac{a}{2}} \cdot 2 \cos \left(2 x\right)\right]$
The derivative of ${b}^{u}$, where b is a number and u is basically anything is: ${b}^{u} \cdot \ln \left(b\right) \cdot u '$, 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 ${5}^{- \frac{a}{2}} \cdot \ln \left(5\right) \cdot - \frac{1}{2}$, and use chain rule to derive the second to get $2 \cos \left(2 x\right)$. Plug these into the product rule and you get the answer, which is probably able to be simplified further..