chain rule: problem 1 (1 point) results for this submission 1 of the answers is not correct. let…

Explanation:

Step1: Recall chain - rule formula

The chain - rule states that if (y = f(u)) and (u = g(x)), then (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). For (y = f(x)=\sqrt{5x^{2}+4x + 2}=(5x^{2}+4x + 2)^{\frac{1}{2}}), let (u = 5x^{2}+4x + 2), so (y = u^{\frac{1}{2}}).

Step2: Differentiate (y) with respect to (u) and (u) with respect to (x)

First, (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}) (using the power - rule (\frac{d}{du}(u^{n})=nu^{n - 1}) with (n=\frac{1}{2})). Second, (\frac{du}{dx}=10x + 4) (using the power - rule (\frac{d}{dx}(ax^{n})=nax^{n - 1}) for each term of (u)).

Step3: Apply the chain - rule

(f^{\prime}(x)=\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2}(5x^{2}+4x + 2)^{-\frac{1}{2}}\cdot(10x + 4)=\frac{10x + 4}{2\sqrt{5x^{2}+4x + 2}}=\frac{5x+2}{\sqrt{5x^{2}+4x + 2}}).

Step4: Evaluate (f^{\prime}(x)) at (x = 2)

Substitute (x = 2) into (f^{\prime}(x)). When (x = 2), (5x^{2}+4x + 2=5\times2^{2}+4\times2 + 2=20 + 8+2=30), and (5x + 2=5\times2+2 = 12). So (f^{\prime}(2)=\frac{5\times2 + 2}{\sqrt{5\times2^{2}+4\times2 + 2}}=\frac{12}{\sqrt{30}}=\frac{12}{\sqrt{6\times5}}=\frac{12}{\sqrt{6}\times\sqrt{5}}=\frac{12\sqrt{30}}{30}=\frac{4\sqrt{30}}{10}=\frac{2\sqrt{30}}{5}\neq\frac{4\sqrt{22}}{11}). The correct derivative evaluation: [ \begin{align*} f^{\prime}(x)&=\frac{5x + 2}{\sqrt{5x^{2}+4x + 2}}\ f^{\prime}(2)&=\frac{5\times2+2}{\sqrt{5\times2^{2}+4\times2 + 2}}=\frac{10 + 2}{\sqrt{20+8 + 2}}=\frac{12}{\sqrt{30}}=\frac{2\sqrt{30}}{5} \end{align*} ]

Answer:

The correct derivative (f^{\prime}(x)=\frac{5x + 2}{\sqrt{5x^{2}+4x + 2}}) and (f^{\prime}(2)=\frac{12}{\sqrt{30}}=\frac{2\sqrt{30}}{5})