How do you use implicit differentiation to find dy/dx given $tanx+csc2y=1$ ?

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Answer 1 · 2017-04-09T11:32:59

$dy/dx=(sec^2 x)/(2csc 2y cot 2y)$

$tan x + csc 2y=1$

By differentiating with respect to $x$ ,

$sec^2 x-csc 2y cot 2y cdot 2 dy/dx=0$

By subtracting $sec^2 x$ from both sides,

$-2csc 2y cot 2y dy/dx=-sec^2 x$

By dividing both sides by $-2csc 2y cot 2y$ ,

$dy/dx=(sec^2 x)/(2csc 2y cot 2y)$

I hope that this was clear.

How do you use implicit differentiation to find dy/dx given tanx+csc2y=1tanx+csc2y=1?