Question

chain rule: problem 1 (1 point) let $f(x)=sqrt{5x^{2}+3x + 3}$ $f(x)=square$ $f(1)=square$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{5x^{2}+3x + 3}=(5x^{2}+3x + 3)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u = 5x^{2}+3x + 3$, 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 $\frac{du}{dx}$: $\frac{du}{dx}=10x + 3$. Then $f^\prime(x)=\frac{1}{2}(5x^{2}+3x + 3)^{-\frac{1}{2}}\cdot(10x + 3)=\frac{10x + 3}{2\sqrt{5x^{2}+3x + 3}}$.

Step3: Evaluate $f^\prime(1)$

Substitute $x = 1$ into $f^\prime(x)$. $f^\prime(1)=\frac{10\times1+ 3}{2\sqrt{5\times1^{2}+3\times1 + 3}}=\frac{13}{2\sqrt{5 + 3+3}}=\frac{13}{2\sqrt{11}}$.

Answer:

$f^\prime(x)=\frac{10x + 3}{2\sqrt{5x^{2}+3x + 3}}$, $f^\prime(1)=\frac{13}{2\sqrt{11}}$