differentiate. f(x)=e^x^2 + 7x f(x)=

Explanation:

Step1: Apply sum - rule of differentiation

The sum - rule states that if (y = u + v), then (y^\prime=u^\prime + v^\prime). Let (u = e^{x^{2}}) and (v = 7x). So (f^\prime(x)=\frac{d}{dx}(e^{x^{2}})+\frac{d}{dx}(7x)).

Step2: Differentiate (v = 7x)

Using the power - rule (\frac{d}{dx}(ax)=a) where (a = 7), we have (\frac{d}{dx}(7x)=7).

Step3: Differentiate (u = e^{x^{2}}) using the chain - rule

Let (t=x^{2}), then (u = e^{t}). By the chain - rule (\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}). We know that (\frac{du}{dt}=e^{t}) and (\frac{dt}{dx}=2x). Substituting (t = x^{2}) back in, (\frac{d}{dx}(e^{x^{2}})=e^{x^{2}}\cdot2x = 2xe^{x^{2}}).

Step4: Combine the results

(f^\prime(x)=2xe^{x^{2}}+7).

Answer:

(2xe^{x^{2}} + 7)