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

1 Answer
Jun 21, 2016

#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)} } #