consider the following curve.\n\n$y = \\sqrt{5 + x^{3}}$\n\nfind $y(x)$.\n\n$y(x)=$\n\nfind an equation of…

Explanation:

Step1: Rewrite the function

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

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u = 5 + x^{3}$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. First, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=3x^{2}$. Substituting $u = 5 + x^{3}$ back in, we get $y^\prime(x)=\frac{1}{2}(5 + x^{3})^{-\frac{1}{2}}\cdot3x^{2}=\frac{3x^{2}}{2\sqrt{5 + x^{3}}}$.

Step3: Find the slope of the tangent line at $x=-1$

Substitute $x = - 1$ into $y^\prime(x)$. $y^\prime(-1)=\frac{3(-1)^{2}}{2\sqrt{5+(-1)^{3}}}=\frac{3}{2\sqrt{4}}=\frac{3}{4}$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-1,2)$ and $m=\frac{3}{4}$. So $y - 2=\frac{3}{4}(x + 1)$.

Step5: Rearrange to slope - intercept form

$y-2=\frac{3}{4}x+\frac{3}{4}$, then $y=\frac{3}{4}x+\frac{3}{4}+2=\frac{3}{4}x+\frac{3 + 8}{4}=\frac{3}{4}x+\frac{11}{4}$.

Answer:

$y^\prime(x)=\frac{3x^{2}}{2\sqrt{5 + x^{3}}}$; $y=\frac{3}{4}x+\frac{11}{4}$