음함수의 미분법

First we need to find the slope of the line tangent to the curve, or $\frac{dy}{dx}$. To do so, we differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ to obtain

$$\begin{aligned}\frac{d}{dx} \left( \sin(x+y) + \cos(x-y) \right) &= \frac{d}{dx}(1) \\ \cos(x+y) \left(1+\frac{dy}{dx}\right) - \sin(x-y)\left(1-\frac{dy}{dx}\right) &= 0\end{aligned}\,.$$ Distributing on the left side of the equation,

$$\cos(x+y) + \cos(x+y)\frac{dy}{dx} - \sin(x-y) + \sin(x-y)\frac{dy}{dx} = 0\,.$$ Removing a factor of $\frac{dy}{dx}$ from two terms on the right and moving the other two terms to the righthand side,

$$\frac{dy}{dx} \left(\cos(x+y) + \sin(x-y) \right) = \sin(x-y) - \cos(x+y)\,.$$ Therefore,

$$\frac{dy}{dx} = \frac{\sin(x-y) - \cos(x+y)}{\cos(x+y) + \sin(x-y)}\,.$$ At the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ we have that

$$\frac{dy}{dx}\biggm|_{\left(\frac{\pi}{2}, \frac{\pi}{2}\right)} = \frac{\sin(0)-\cos(\pi)}{\cos(\pi) + \sin(0)} = -1\,.$$ Therefore, the equation of the line tangent to the curve at the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ is $y = \frac{\pi}{2} - \left(x-\frac{\pi}{2}\right)$. A portion of the graph of the curve defined by $\sin(x+y) + \cos(x-y) = 1$ and the tangent line to the curve at the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ are shown in the following figure.