How do you find the derivative of #g(alpha)=5^(alpha/2)sin2alpha#?
You could probably simplify this more though
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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:

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} ).

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.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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