Question

find the derivative of ( f(x) ).

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

( 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:

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