find the derivative of ( f(x) ).\n\n( f(x)=3^{x}+e^{-x} )\n\n( f^{prime}(x)= )

Explanation:

Step1: Differentiate (3^x)

The derivative of (a^x) is (a^x\ln a). So, the derivative of (3^x) is (3^x\ln 3).

Step2: Differentiate (e^{-x})

Using the chain rule, if (y = e^{u}) and (u=-x), then (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). The derivative of (e^{u}) with respect to (u) is (e^{u}), and the derivative of (u = -x) with respect to (x) is (-1). So, the derivative of (e^{-x}) is (e^{-x}\cdot(- 1)=-e^{-x}).

Step3: Combine the derivatives

Since (f(x)=3^x + e^{-x}), by the sum rule ((u + v)'=u'+v'), we have (f'(x)=(3^x)'+(e^{-x})').

Answer:

(3^x\ln 3 - e^{-x})