(1 point)\nlet\n$f(x)=\\sqrt{5 x^{2}+2 x+2}$\n$f^{prime}(x)=\\square$\n$f^{prime}(1)=\\square$\nnote: you…

Explanation:

Step1: Use the chain rule

Let (u = 5x^{2}+2x + 2), then (y=\sqrt{u}=u^{\frac{1}{2}}). The chain rule states that (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). First, find (\frac{dy}{du}): (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}). Second, find (\frac{du}{dx}): (\frac{du}{dx}=10x + 2). Then (f^{\prime}(x)=\frac{1}{2}(5x^{2}+2x + 2)^{-\frac{1}{2}}\cdot(10x + 2)=\frac{10x + 2}{2\sqrt{5x^{2}+2x + 2}}=\frac{5x + 1}{\sqrt{5x^{2}+2x + 2}}).

Step2: Substitute (x = 1) into (f^{\prime}(x))

When (x = 1), (f^{\prime}(1)=\frac{5\times1+1}{\sqrt{5\times1^{2}+2\times1+2}}=\frac{6}{\sqrt{5 + 2+2}}=\frac{6}{3}=2).

Answer:

(f^{\prime}(x)=\frac{5x + 1}{\sqrt{5x^{2}+2x + 2}}) (f^{\prime}(1)=2)