Help with derivative using Implicit Differentiation

Question

I need help solving this problem

$\mathbf{y=\frac{1}{3}xy^2 - \frac{1}{4}ysin12x}$

This is my solution for this problem but I'm not quite certain of it.

First, I derived the equation

$\frac{dy}{dx} = [(\frac{1}{3}\bullet y^2)+(2y\frac{dy}{dx}\bullet \frac{1}{3}x)] - [(\frac{1}{4}\frac{dy}{dx}\bullet sin12x)+(12cos12x\bullet\frac{1}{4}y)]$

Then I isolated the $\frac{dy}{dx}$

$\frac{dy}{dx}-2y\frac{dy}{dx}\bullet \frac{1}{3}x+\frac{1}{4}\frac{dy}{dx}\bullet sin12x = \frac{1}{3}y^2-(12cos12x\bullet \frac{1}{4}y)$

factor out $\frac{dy}{dx}$

$\frac{dy}{dx}(1-2y\frac{1}{3}x+\frac{1}{4}sin12x) = \frac{1}{3}y^2-(12cos12x\bullet \frac{1}{4}y)$

Solving for $\frac{dy}{dx}$:

$\frac{\frac{dy}{dx}(1-2y\frac{1}{3}x+\frac{1}{4}sin12x)}{(1-2y\frac{1}{3}x+\frac{1}{4}sin12x)} = \frac{\frac{1}{3}y^2-(12cos12x\bullet \frac{1}{4}y)}{(1-2y\frac{1}{3}x+\frac{1}{4}sin12x)}$

thus,

$\frac{dy}{dx}=\frac{\frac{1}{3}y^2-12cos12x\frac{1}{4}y}{1-2y\frac{1}{3}x+\frac{1}{4}sin12x}$

I need someone to verify this for me, any help would be greatly appreciated!

Answer 1

Yes, your result is correct, although a bit excessively computational due to not simplifying intermediate results in your calculation. You may find the following a bit easier:

$$\begin{align*} y &= \frac{xy^2}{3} - \frac{y \sin 12x}{4} \\ 12y &= 4xy^2 - 3y \sin 12x \\ 3y(4+\sin 12 x) &= 4xy^2 \\ 3(4+\sin 12x) &= 4xy \\ y &= \frac{3(4+\sin 12x)}{4x}. \end{align*}$$

Then differentiating explicitly gives the same result when you substitute for $y$ using the last expression above.

If you require implicit differentiation of the original expression, we might write $$\begin{align*} \frac{dy}{dx} &= \frac{1}{3}\left(y^2 + 2xy \frac{dy}{dx} \right) - \frac{1}{4}\left(\frac{dy}{dx} \sin 12x + 12 y \cos 12x \right) \\ \frac{dy}{dx}\left(1 - \frac{2xy}{3} + \frac{\sin 12x}{4} \right) &= \frac{y^2}{3} - 3y \cos 12x \\ \frac{dy}{dx} &= \frac{y^2 - 9y \cos 12x }{3 - 2xy + \frac{3}{4} \sin 12x},\end{align*}$$ where upon substituting $y = 3(4+\sin 12x)/(4x)$ we get $$\frac{dy}{dx} = \frac{3(12x \cos 12x - \sin 12x - 4)}{4x^2}, \quad y \ne 0.$$

The full plot of the given relation includes the line $y = 0$, but if we omit this line, we get a legitimate function defined on all $x \ne 0$:

Answer 2

i got this here $$y'=\frac{1}{3}y^2+\frac{1}{3}x\cdot 2y\cdot y'-\frac{1}{4}y'\sin(12x)-\frac{1}{4}y\cos(12x)\cdot 12$$ solving for $y'$ we get $$y'=\frac{\frac{1}{3}y^2-3y\cos(12x)}{1-\frac{2}{3}xy+\frac{1}{4}\sin(12x)}$$