Answer
$$y=2x-1$$
Work Step by Step
Finding the derivative of the curve:
$y'=\frac{d\sqrt{1+x^3}}{dx}$
Using the chain rule:
$y'=\frac{d\sqrt{1+x^3}}{d(1+x^3)}
\times \frac{d(1+x^3)}{dx}$
$=\frac{1}{2\sqrt{1+x^3}}
\times 3x^2$
$=\frac{3x^2}{2\sqrt{1+x^3}}$
Finding the slope ($m$) of the tangent line at (2,3)
$m=y'(2)$
$m=\frac{3(2)^2}{2\sqrt{1+(2)^3}}=\frac{12}{2\sqrt{9}}=2$
From the general point-slope form:
$y-3=2(x-2)$
Thus, an equation of the tangent line would be:
$y=2x-4+3=2x-1$
