Question

find the derivative of ( f(x) ).

( f(x)=e^{-x}-2^{x} )

( f^{prime}(x)= )

Explanation:

Step1: Differentiate \(e^{-x}\)

Use the chain rule. Let \(u = -x\), then \(\frac{d}{dx}(e^u)=e^u\cdot\frac{du}{dx}\). Here \(\frac{du}{dx}=- 1\), so \(\frac{d}{dx}(e^{-x})=e^{-x}\cdot(-1)=-e^{-x}\)

Step2: Differentiate \(2^{x}\)

Use the formula \(\frac{d}{dx}(a^{x})=a^{x}\ln a\). For \(a = 2\), \(\frac{d}{dx}(2^{x})=2^{x}\ln2\)

Step3: Combine the derivatives

Since \(f(x)=e^{-x}-2^{x}\), by the sum - difference rule \((u - v)'=u'-v'\), where \(u = e^{-x}\) and \(v = 2^{x}\). So \(f'(x)=\frac{d}{dx}(e^{-x})-\frac{d}{dx}(2^{x})\)

Answer:

\(-e^{-x}-2^{x}\ln2\)