find the derivative of ( f(x) ).\n\n( f(x)=4^{x}+e^{-3 x} )\n\n( f^{prime}(x)= )

Explanation:

Step1: Differentiate (4^{x})

Use the formula (\frac{d}{dx}(a^{x})=a^{x}\ln a). For (a = 4), (\frac{d}{dx}(4^{x})=4^{x}\ln 4).

Step2: Differentiate (e^{-3x})

Use the chain - rule (\frac{d}{dx}(e^{u})=e^{u}\cdot u'). Let (u=-3x), then (u'=-3). So (\frac{d}{dx}(e^{-3x})=e^{-3x}\cdot(-3)=-3e^{-3x}).

Step3: Combine the derivatives

Since (f(x)=4^{x}+e^{-3x}), by the sum - rule ((u + v)'=u'+v'), (f'(x)=\frac{d}{dx}(4^{x})+\frac{d}{dx}(e^{-3x})).

Answer:

(f'(x)=4^{x}\ln 4-3e^{-3x})