How do you find the derivative of $tan 2x = cos 3y$ ?

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Answer 1 · 2018-03-16T14:09:12

$dy/dx = -(2sec^2 2x)/(3 sin 3y)$

Differentiate both sides of the equation with respect to $x$ :

$d/dx (tan2x) = d/dx (cos3y)$

using the chain rule:

$2sec^2 2x = -3dy/dxsin3y$

$dy/dx = -(2sec^2 2x)/(3 sin 3y)$

How do you find the derivative of tan 2x = cos 3y tan 2x = cos 3y?