Question

find the derivative of ( f(x) ).

( f(x)=4^{x}+e^{-3 x} )

( 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:

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