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

Question

Ezra Adams · Accepted Answer

#-(2sec^2(2x)*csc(3y))/3 = dy/dx# Using the chain rule, where #u = 2x# and #d/dx = d/(du)*(du)/dx# #d/(du)(tan(u))*(du)/dx = d/dx(cos(3y))# #sec^2(u)*2 = d/dx(cos(3y))# #2sec^2(2x) = d/dx(cos(3y))# Using the chain rule, where #v = 3y# and #d/dx = d/(dv)*(dv)/dy*dy/dx# #2sec^2(2x) = d/(dv)(cos(v))*d/dy(3y)*dy/dx# #2sec^2(2x) = -3sin(v)*dy/dx# #-(2sec^2(2x)*csc(3y))/3 = dy/dx#

Christopher Agresta · Answer

undefined To differentiate $ \tan(2x) = \cos(3y) $, you can use implicit differentiation. The derivative of both sides of the equation with respect to $ x $ is taken.

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

$$ \frac{d}{dx}[\tan(2x)] = \frac{d}{dx}[\cos(3y)] $$

2. Use the chain rule and trigonometric derivative rules:

$$ \sec^2(2x) \cdot 2 = -\sin(3y) \cdot \frac{d}{dx}(3y) $$

3. Solve for $ \frac{dy}{dx} $ by isolating $ \frac{d}{dx}(3y) $:

$$ \frac{d}{dx}(3y) = \frac{2}{\sec^2(2x)} \cdot -\sin(3y) $$

$$ \frac{d}{dx}(3y) = -2\sin(3y)\sec^2(2x) $$

4. Finally, use $ \frac{dy}{dx} = \frac{1}{3} \cdot \frac{d}{dx}(3y) $ to find $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} = -\frac{2}{3}\sin(3y)\sec^2(2x) $$