Question

let $f(x)=sqrt{3x^{2}+5x + 8}$
$f(x)=$
enter an algebraic expression more...
$f(1)=$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{3x^{2}+5x + 8}=(3x^{2}+5x + 8)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u = 3x^{2}+5x + 8$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{d}{du}(u^{\frac{1}{2}})=\frac{1}{2}u^{-\frac{1}{2}}$. Second, find $\frac{du}{dx}$: $\frac{d}{dx}(3x^{2}+5x + 8)=6x + 5$. Then $f^{\prime}(x)=\frac{1}{2}(3x^{2}+5x + 8)^{-\frac{1}{2}}\cdot(6x + 5)=\frac{6x + 5}{2\sqrt{3x^{2}+5x + 8}}$.

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

Substitute $x = 1$ into $f^{\prime}(x)$: $f^{\prime}(1)=\frac{6\times1+5}{2\sqrt{3\times1^{2}+5\times1 + 8}}=\frac{6 + 5}{2\sqrt{3+5 + 8}}=\frac{11}{2\sqrt{16}}=\frac{11}{2\times4}=\frac{11}{8}$.

Answer:

$f^{\prime}(x)=\frac{6x + 5}{2\sqrt{3x^{2}+5x + 8}}$
$f^{\prime}(1)=\frac{11}{8}$