15–48. Derivatives Find the derivative of the following functions.
y = 4^-x sin x

Verified step by step guidance

1. Step 1: Identify the function as a product of two functions, $y = 4^{-x} \sin x$. This suggests using the product rule for differentiation.
2. Step 2: Recall the product rule for derivatives, which states that if $y = u(x) \cdot v(x)$, then $y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$. Here, let $u(x) = 4^{-x}$ and $v(x) = \sin x$.
3. Step 3: Differentiate $u(x) = 4^{-x}$. Use the chain rule: $u'(x) = \frac{d}{dx}(4^{-x}) = 4^{-x} \cdot \ln(4) \cdot (-1)$, which simplifies to $-4^{-x} \ln(4)$.
4. Step 4: Differentiate $v(x) = \sin x$. The derivative is straightforward: $v'(x) = \cos x$.
5. Step 5: Apply the product rule: $y' = (-4^{-x} \ln(4)) \cdot \sin x + 4^{-x} \cdot \cos x$. This expression represents the derivative of the given function.