Calculus

Problem:

Solution:

Implicitly differentiate both side, we have

$ \frac{d(\cot{2x} - 2\csc{xy})}{dx} = \frac{dy^2}{dx} $
$ -2\csc{2x} + 2\csc(xy)\cot(xy)(y + x\frac{dy}{dx})= 2y\frac{dy}{dx} $
$ -2\csc{2x} + 2y\csc(xy)\cot(xy)= (2y - 2x\csc(xy)\cot(xy))\frac{dy}{dx} $
$ \frac{dy}{dx} = \frac{2y\csc(xy)\cot(xy) - 2\csc{2x}}{2y - 2x\csc(xy)\cot(xy)} $

The next problem is simply repeated application of the chain rule:

$ \frac{dy}{dx} = 3(3e^{4x} + 4x^2\ln(x))\frac{d(3e^{4x} + 4x^2\ln(x))}{dx} $
$ \frac{dy}{dx} = 3(3e^{4x} + 4x^2\ln(x))(12e^{4x} + \frac{d(4x^2\ln(x))}{dx}) $
$ \frac{dy}{dx} = 3(3e^{4x} + 4x^2\ln(x))(12e^{4x} + 4x^2\frac{1}{x} + 8x\ln(x)) $
$ \frac{dy}{dx} = 3(3e^{4x} + 4x^2\ln(x))(12e^{4x} + 4x + 8x\ln(x)) $

The last problem is trivial, just plot it.