Question: Find the slope of the tangent line to the inverse of $x^2 = \cos^{-1}(xy)$ at the point $(\frac{1}{3}, 2)$. Note that $y = f(x)$ and truncate your answer to two decimal places.
1. **State the problem:**
Find the slope of the tangent line to the inverse of the curve defined implicitly by $$x^2 = \cos^{-1}(xy)$$ at the point $$\left(\frac{1}{3}, 2\right)$$.
2. **Understand the problem:**
We have an implicit relation involving $x$ and $y$. We want the slope of the tangent line to the inverse function, which means if the original function is $y=f(x)$, the inverse function is $x=f^{-1}(y)$. The slope of the tangent line to the inverse at a point is the reciprocal of the slope of the original function at the corresponding point.
3. **Find $\frac{dy}{dx}$ using implicit differentiation:**
Given:
$$x^2 = \cos^{-1}(xy)$$
Differentiate both sides with respect to $x$:
$$\frac{d}{dx}(x^2) = \frac{d}{dx}\left(\cos^{-1}(xy)\right)$$
Left side:
$$\frac{d}{dx}(x^2) = 2x$$
Right side:
Using chain rule:
$$\frac{d}{dx}\left(\cos^{-1}(u)\right) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$
where $$u = xy$$.
Calculate $$\frac{du}{dx}$$:
$$\frac{d}{dx}(xy) = y + x \frac{dy}{dx}$$
So,
$$\frac{d}{dx}\left(\cos^{-1}(xy)\right) = -\frac{1}{\sqrt{1-(xy)^2}} (y + x \frac{dy}{dx})$$
4. **Set up the equation:**
$$2x = -\frac{1}{\sqrt{1-(xy)^2}} (y + x \frac{dy}{dx})$$
Multiply both sides by $$\sqrt{1-(xy)^2}$$:
$$2x \sqrt{1-(xy)^2} = - (y + x \frac{dy}{dx})$$
Rearranged:
$$y + x \frac{dy}{dx} = -2x \sqrt{1-(xy)^2}$$
5. **Solve for $$\frac{dy}{dx}$$:**
$$x \frac{dy}{dx} = -2x \sqrt{1-(xy)^2} - y$$
Divide both sides by $$x$$:
$$\frac{dy}{dx} = -2 \sqrt{1-(xy)^2} - \frac{y}{x}$$
6. **Evaluate at the point $$\left(\frac{1}{3}, 2\right)$$:**
Calculate $$xy$$:
$$xy = \frac{1}{3} \times 2 = \frac{2}{3}$$
Calculate $$\sqrt{1-(xy)^2}$$:
$$\sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$$
Substitute values:
$$\frac{dy}{dx} = -2 \times \frac{\sqrt{5}}{3} - \frac{2}{\frac{1}{3}} = -\frac{2\sqrt{5}}{3} - 6$$
7. **Calculate numeric value:**
$$\sqrt{5} \approx 2.236$$
So,
$$\frac{dy}{dx} \approx -\frac{2 \times 2.236}{3} - 6 = -\frac{4.472}{3} - 6 = -1.49 - 6 = -7.49$$
8. **Find slope of tangent to inverse function:**
The slope of the inverse function at the point is the reciprocal:
$$m_{inverse} = \frac{1}{\frac{dy}{dx}} = \frac{1}{-7.49} \approx -0.13$$
**Final answer:**
$$\boxed{-0.13}$$