Question

Find the derivative of the function. \(g(\alpha)=5^{-\alpha / 2} \sin 2 \alpha\)

Short answer

The derivative of the function \(g(\alpha)=5^{-\alpha / 2} \sin 2 \alpha\) is \( g'(\alpha) = 5^{-\alpha/2}(2\cos(2\alpha) - \ln 5\sin(2\alpha)/2) \).

Step-by-step solution

Identify the inner function

Identify the inner function in the exponent of the base 5, which is \(-\alpha/2\). Let's denote it as \( u(\alpha) = -\alpha/2 \).

Identify the outer function

Identify the outer function which is the sine function, we'll call it \( v(\alpha) = \sin(2\alpha) \). Our function g then becomes \( g(\alpha) = 5^{u(\alpha)}v(\alpha) = e^{u(\alpha)\ln 5}v(\alpha)\).

Apply the Product Rule

Because our function g is a product of two functions, we can apply the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Denoting g' as the derivative of g, the product rule gives us \( g'(\alpha) = e^{u(\alpha)\ln 5}v'(\alpha) + (e^{u(\alpha)\ln 5})'v(\alpha) \).

Compute Derivatives and Substitute

Now, we compute the derivatives in the equation above. The derivative of the sine function \(v'(\alpha) = 2\cos(2\alpha)\), while for the exponential part we apply the chain rule, giving us \((e^{u(\alpha)\ln 5})' = e^{u(\alpha)\ln 5}u'(\alpha)\ln 5\), where \(u'(\alpha) = -1/2\). Substituting these gives us \(g'(\alpha) = e^{u(\alpha)\ln 5}2\cos(2\alpha) - e^{u(\alpha)\ln 5}/2\ln 5\sin(2\alpha)\).

Simplify the Result

After substituting into the formula, simplify the result to get the final derivative of \( g(\alpha) \). This gives us \( g'(\alpha) = 5^{-\alpha/2}(2\cos(2\alpha) - \ln 5\sin(2\alpha)/2) \).