Question

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

Answer 1

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

Explanation:

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)`