find the derivative of the function.\ny = 5e^{-x}+e^{4x}\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Apply sum - rule of derivatives

The derivative of a sum of functions (y = u + v) is (\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}). Let (u = 5e^{-x}) and (v=e^{4x}). So (\frac{dy}{dx}=\frac{d(5e^{-x})}{dx}+\frac{d(e^{4x})}{dx}).

Step2: Find the derivative of (u = 5e^{-x})

Using the constant - multiple rule (\frac{d(cf(x))}{dx}=c\frac{df(x)}{dx}) and the chain - rule (\frac{d(e^{ax})}{dx}=ae^{ax}), for (u = 5e^{-x}), we have (\frac{d(5e^{-x})}{dx}=5\frac{d(e^{-x})}{dx}). Since (\frac{d(e^{-x})}{dx}=-e^{-x}), then (\frac{d(5e^{-x})}{dx}=5\times(-e^{-x})=- 5e^{-x}).

Step3: Find the derivative of (v = e^{4x})

Using the chain - rule (\frac{d(e^{ax})}{dx}=ae^{ax}), for (v = e^{4x}) with (a = 4), we get (\frac{d(e^{4x})}{dx}=4e^{4x}).

Step4: Combine the results

(\frac{dy}{dx}=-5e^{-x}+4e^{4x}).

Answer:

(-5e^{-x}+4e^{4x})