determine the slope of the line tangent to the curve $xcos(y)=ysin(x)$ at $(\\frac{\\pi}{4},\\frac{\\pi}{4})$.

Explanation:

Step1: Differentiate both sides implicitly

Differentiate $x\cos(y)=y\sin(x)$ with respect to $x$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, we have: $\cos(y)-x\sin(y)y^\prime=y\cos(x)+\sin(x)y^\prime$.

Step2: Solve for $y^\prime$

Rearrange the terms to isolate $y^\prime$: $-x\sin(y)y^\prime-\sin(x)y^\prime=y\cos(x)-\cos(y)$. Factor out $y^\prime$: $y^\prime(-x\sin(y)-\sin(x))=y\cos(x)-\cos(y)$. So, $y^\prime=\frac{y\cos(x)-\cos(y)}{-x\sin(y)-\sin(x)}$.

Step3: Substitute the point $(\frac{\pi}{4},\frac{\pi}{4})$

Substitute $x = \frac{\pi}{4}$ and $y=\frac{\pi}{4}$ into $y^\prime$: [ \begin{align*} y^\prime&=\frac{\frac{\pi}{4}\cos(\frac{\pi}{4})-\cos(\frac{\pi}{4})}{-\frac{\pi}{4}\sin(\frac{\pi}{4})-\sin(\frac{\pi}{4})}\ &=\frac{\frac{\pi}{4}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}}{-\frac{\pi}{4}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}}\ &=\frac{\frac{\sqrt{2}}{2}(\frac{\pi}{4} - 1)}{-\frac{\sqrt{2}}{2}(\frac{\pi}{4}+1)}\ &=\frac{1-\frac{\pi}{4}}{1 + \frac{\pi}{4}}\ &=\frac{4 - \pi}{4+\pi} \end{align*} ]

Answer:

$\frac{4 - \pi}{4+\pi}$