Question

How do you differentiate `tan 2x = cos 3y`?

Answer 1

`y' = -\frac{2 sec^2 (2x)}{3 sqrt{1 - tan^2(2x)} }`

Explanation:

`\frac{d}{dx} \tan(2x) = 2 sec^2 (2x)`

if you are uncomfortable with this, make the sub `u = 2x` and use the chain rule so `\frac{d}{dx} \tan(2x) =\frac{d}{dx} \tan(u(x)) = \frac{d}{du} \tan(u) \frac{du}{dx} = sec^2 u (2) = 2 sec^2 (2x)`

similarly: `\frac{d}{dx} \cos(3y) = -3 sin (3y) \ y'`

`\implies y' = -\frac{2 sec^2 (2x)}{3 sin (3y)}`

using the identity `cos^2 + sin^2 = 1` we can say that `3sin(3y) = 3 sqrt{1 - cos^2(3y)} = 3 sqrt{1 - tan^2(2x)}`

so `y' = -\frac{2 sec^2 (2x)}{3 sqrt{1 - tan^2(2x)} }`