Question

chain rule: problem 1 (1 point) let ( f(x)=sqrt{3x^{2}+4x + 2}) ( f(x)=square) ( f(2)=square) note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{3x^{2}+4x + 2}=(3x^{2}+4x + 2)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u = 3x^{2}+4x + 2$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$. Second, find $\frac{du}{dx}$: $\frac{du}{dx}=6x + 4$. Then $f^\prime(x)=\frac{1}{2}(3x^{2}+4x + 2)^{-\frac{1}{2}}\cdot(6x + 4)=\frac{6x + 4}{2\sqrt{3x^{2}+4x + 2}}=\frac{3x + 2}{\sqrt{3x^{2}+4x + 2}}$.

Step3: Evaluate $f^\prime(2)$

Substitute $x = 2$ into $f^\prime(x)$: $f^\prime(2)=\frac{3\times2+2}{\sqrt{3\times2^{2}+4\times2 + 2}}=\frac{6 + 2}{\sqrt{12 + 8+2}}=\frac{8}{\sqrt{22}}=\frac{8\sqrt{22}}{22}=\frac{4\sqrt{22}}{11}$.

Answer:

$f^\prime(x)=\frac{3x + 2}{\sqrt{3x^{2}+4x + 2}}$; $f^\prime(2)=\frac{4\sqrt{22}}{11}$