find the derivative of f(x).\nf(x) = -2·7^x - e^x\nf(x) =

Explanation:

Step1: Differentiate (-2\cdot7^{x})

Use the formula (\frac{d}{dx}(a^{x}) = a^{x}\ln a). Here (a = 7), so (\frac{d}{dx}(-2\cdot7^{x})=-2\cdot7^{x}\ln7)

Step2: Differentiate (-e^{x})

Use the formula (\frac{d}{dx}(e^{x}) = e^{x}). So (\frac{d}{dx}(-e^{x})=-e^{x})

Step3: Combine the derivatives

By the sum - rule of differentiation ((u + v)^\prime=u^\prime + v^\prime), where (u=-2\cdot7^{x}) and (v = -e^{x}). Then (f^\prime(x)=\frac{d}{dx}(-2\cdot7^{x})+\frac{d}{dx}(-e^{x}))

Answer:

(-2\cdot7^{x}\ln7 - e^{x})