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Given xsin(2y)=ycos(2x), find dy/dx at x=pi/4
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\[\Large\bf\sf x \sin(2y)\quad=\quad y \cos(2x)\]So we'll want to differentiate implicitly (without separating the y's and x's). Can you see the equation I wrote above? Or does it just look like a big mess of code?
The LaTeX plugin hasn't been working properly the last couple of days.. just wanted to see if we can use it now :O
x cos(2y)(dy/dx)+1*sin(2y)=y[-sin(2x)*2]+cos(2x)(dy/dx) [xcos(2y)-cos(2x)]dy/dx=-2ysin(2x)-sin(2y) dy/dx={-2y sin(2x)-sin(2y)}/{x cos (2y)-cos (2x)}
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