## Calculus: Early Transcendentals 8th Edition

$$y=2x-1$$
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: $y=2x-4+3=2x-1$