Find the equation of the tangent to the curve $y = \frac{\sqrt{x}-2}{\sqrt[3]{x}}$ at P (1, -1).

Steps to Solve

1. Rewrite the function in terms of exponents Rewrite the function $y = \frac{\sqrt{x}-2}{\sqrt[3]{x}}$ using exponents instead of radicals: $y = \frac{x^{1/2}-2}{x^{1/3}}$

2. Apply the quotient rule Differentiate $y$ with respect to $x$ using the quotient rule: $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, $u = x^{1/2} - 2$ and $v = x^{1/3}$.

$\frac{du}{dx} = \frac{1}{2}x^{-1/2}$ $\frac{dv}{dx} = \frac{1}{3}x^{-2/3}$

Now, apply the quotient rule: $\frac{dy}{dx} = \frac{x^{1/3}(\frac{1}{2}x^{-1/2}) - (x^{1/2}-2)(\frac{1}{3}x^{-2/3})}{(x^{1/3})^2}$ $\frac{dy}{dx} = \frac{\frac{1}{2}x^{-1/6} - \frac{1}{3}x^{-1/6} + \frac{2}{3}x^{-2/3}}{x^{2/3}}$

3. Simplify the derivative Simplify the expression: $\frac{dy}{dx} = \frac{\frac{1}{6}x^{-1/6} + \frac{2}{3}x^{-2/3}}{x^{2/3}}$ $\frac{dy}{dx} = \frac{1}{6}x^{-1/6 - 2/3} + \frac{2}{3}x^{-2/3 - 2/3}$ $\frac{dy}{dx} = \frac{1}{6}x^{-5/6} + \frac{2}{3}x^{-4/3}$

4. Evaluate the derivative at $x=1$ Substitute $x=1$ into the derivative to find the slope of the tangent at the point $(1, -1)$: $m = \frac{dy}{dx}\Big|_{x=1} = \frac{1}{6}(1)^{-5/6} + \frac{2}{3}(1)^{-4/3} = \frac{1}{6} + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}$

5. Find the equation of the tangent line Use the point-slope form of a line: $y - y_1 = m(x - x_1)$, where $(x_1, y_1) = (1, -1)$ and $m = \frac{5}{6}$. $y - (-1) = \frac{5}{6}(x - 1)$ $y + 1 = \frac{5}{6}x - \frac{5}{6}$ $y = \frac{5}{6}x - \frac{5}{6} - 1$ $y = \frac{5}{6}x - \frac{5}{6} - \frac{6}{6}$ $y = \frac{5}{6}x - \frac{11}{6}$