Use implicit differentiation to find an equation of the tangent line to the curve at the given point. Given $y\sin(8x) = x\cos(2y)$ find the tangent line at the point ($\pi\over2$, $\pi\over4$).
I got $y = 2x - 2.36$, but my teacher wants a fraction. Can somebody help me get the answer in fraction form? 
 A: You have $$y\sin(8x)=x\cos(2y)$$
Then
$$y'\sin(8x)+8y \cos(8x)=\cos(2y)-2y'x\sin(2y)$$
Thus
$$y'(\sin(8x)+2x\sin(2y))=\cos(2y)-8y\cos(8x)$$
$$y'=\frac{\cos(2y)-8y\cos(8x)}{\sin(8x)+2x\sin(2y)}$$
We look at $$(\pi/2,\pi/4)$$
$$y'_{(\pi/2,\pi/4)}=\frac{\cos(2 \pi/4)-8 \pi/4\cos(8 \pi/2)}{\sin(8 \pi/2)+2 \pi/2\sin(2 \pi/4)}$$
$$\eqalign{
  & {{y'}_{(\pi /2,\pi /4)}} = \frac{{\cos (\pi /2) - 2\pi \cos (4\pi )}}{{\sin (4\pi ) + \pi \sin (\pi /2)}}  \cr 
  & {{y'}_{(\pi /2,\pi /4)}} = \frac{{0 - 2\pi }}{{0 + \pi }} =  - 2 \cr} $$
Since $y(\pi/2)=\pi/4$, we get
$$y_T=-2(x-\pi/2)+\pi/4$$
$$y_T=-2x+\pi+\pi/4=-2x+5\pi/4$$
A: General and short formula:

If $F(x,y)=0$ defines $y$ as a function of $x$ implicity, then we always have $y'=\frac{-F_x}{F_y}$.

Here we have $y\sin(8x) = x\cos(2y)$ so $$F(x,y)=y\sin(8x) - x\cos(2y)=0$$ and then $$F_x=8y\cos(8x)-\cos(2y),~~~F_y=\sin(8x)+x\sin(2y)$$
A: Differentiate (implicitly). We get
$$8y\cos(8x)+\sin(8x)y'=-2x\sin(2y)y'+\cos(2y).$$
When we substitute the given values of $x$ and $y$, the numbers become very simple, since $\cos 2y=0$ and $\sin 8x=0$.  We get that at our target point,
$$2\pi =-\pi y',$$
and our slope is $-2$.
The tangent line therefore has equation
$$y-\frac{\pi}{4}=-2\left(x-\frac{\pi}{2}\right).$$
This simplifies to $y=-2x +\dfrac{5\pi}{4}.$
A: Differentiating implicitly the equation $\,y\sin 8x=x\cos 2y\,$ ,we get
$$\sin 8x\,dy+8y\cos 8x\,dx=\cos 2y\,dx-2x\sin 2y\,dy\Longrightarrow$$
$$(\sin 8x+2x\sin 2y)dx=(\cos 2y-8y\cos 8x)dx\Longrightarrow$$
$$\frac{dy}{dx}=\frac{\cos 2y-8y\cos 8x}{\sin 8x+2x\sin 2y}\Longrightarrow$$
$$\left.\frac{dy}{dx}\right|_{\left(\frac{\pi}{2},\frac{\pi}{4}\right)}=\frac{0-2\pi}{0+\pi}=-2$$
Thus, the tangent line's given by
$$y-\frac{\pi}{4}=-2\left(x-\frac{\pi}{2}\right)\Longleftrightarrow y=-2x+\frac{5\pi}{4}$$
