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

Question

Charles Anctil · Accepted Answer

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

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..

Christopher Allers · Answer

undefined The user is interested in math and specifically asked about finding the derivative of a function.To find the derivative of $ g(\alpha) = 5^{-\frac{\alpha}{2}} \sin(2\alpha) $, you can use the product rule and chain rule. First, differentiate each part separately:

1. For $ 5^{-\frac{\alpha}{2}} $:
   - Use the chain rule: $ \frac{d}{d\alpha} 5^{-\frac{\alpha}{2}} = -\frac{1}{2} \cdot 5^{-\frac{\alpha}{2}-1} \cdot \frac{d}{d\alpha} \alpha $.
   - Simplify to get $ \frac{d}{d\alpha} 5^{-\frac{\alpha}{2}} = -\frac{1}{2} \cdot 5^{-\frac{\alpha}{2}-1} $.

2. For $ \sin(2\alpha) $:
   - Use the chain rule: $ \frac{d}{d\alpha} \sin(2\alpha) = 2 \cos(2\alpha) $.

Now, apply the product rule: $ \frac{d}{d\alpha} (f(\alpha) \cdot g(\alpha)) = f'(\alpha) \cdot g(\alpha) + f(\alpha) \cdot g'(\alpha) $, where $ f(\alpha) = 5^{-\frac{\alpha}{2}} $ and $ g(\alpha) = \sin(2\alpha) $:

$ g'(\alpha) = \frac{d}{d\alpha} 5^{-\frac{\alpha}{2}} \cdot \sin(2\alpha) + 5^{-\frac{\alpha}{2}} \cdot \frac{d}{d\alpha} \sin(2\alpha) $.

Substitute the derivatives we found earlier to get the final answer:

$ g'(\alpha) = -\frac{1}{2} \cdot 5^{-\frac{\alpha}{2}-1} \cdot \sin(2\alpha) + 5^{-\frac{\alpha}{2}} \cdot 2 \cos(2\alpha) $.

Simplify if needed.