91Ó°ÊÓ

First we will apply the implicit differentiation to the given equation, \(\tan x y = x + y\). We have to differentiate each term separately with respect to x. The derivative of \(\tan x y\) with respect to x can be found using the chain rule: \(\frac{d}{dx}(\tan x y) = \frac{d}{dx} (\tan(u)) \frac{d}{dx}(xy)\), where \(u=xy\). Now, \(\frac{d}{dx}(\tan(u)) = \sec^2(u) \cdot \frac{du}{dx}\), and \(\frac{d}{dx}(xy) = y+x\frac{dy}{dx}\), using the product rule. So the derivative of the left side is: \(\sec^2(xy) (y + x\frac{dy}{dx})\). Now we'll differentiate the right side \((x+y)\) with respect to x: \(\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}\). Now we have: \(\sec^2(xy) (y + x\frac{dy}{dx}) = 1 + \frac{dy}{dx}\).

Now, we need to isolate \(\frac{dy}{dx}\) in the equation we derived: \(\sec^2(xy) (y + x\frac{dy}{dx}) = 1 + \frac{dy}{dx}\). We can start by distributing \(\sec^2(xy)\) to both terms inside the parenthesis: \(\sec^2(xy) y + \sec^2(xy)x\frac{dy}{dx} = 1 + \frac{dy}{dx}\). Next, we can subtract the \(\frac{dy}{dx}\) term from both sides: \(\sec^2(xy) y - \frac{dy}{dx} + \sec^2(xy)x\frac{dy}{dx} = 1\). Now, we can factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(1 - \sec^2(xy)x) = 1 - \sec^2(xy)y\). Finally, we divide by \((1-\sec^2(xy)x)\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{1 - \sec^2 (xy)y}{1 - \sec^2(xy)x}\).

Now that we have the derivative, we can find the slope of the curve at the point (0, 0) by plugging in the x and y values: \(\frac{dy}{dx} \bigg|_{(0,0)} = \frac{1 - \sec^2 (0)(0)}{1 - \sec^2(0)(0)}\). Since \(\sec(0) = 1\), our expression simplifies to: \(\frac{dy}{dx} \bigg|_{(0,0)} = \frac{1 - 1(0)}{1 - 1(0)} = \frac{1}{1}\). So the slope of the curve at the given point (0, 0) is 1.