Use implicit differentiation to find dy/dx for $x \ge -2$. $y\sqrt{x+2}=xy+3$; I am getting a different answer than is given

Question

$$y\sqrt{x+2}=xy+3$$ find $\frac {dy}{dx}$ for $x \ge -2$.

The given answer is $$-\frac{y(2(x+2)^{1/2}-1)}{2(x(x+2)^{1/2}-x-2)}$$

Here's my work so far:

$$y\sqrt{x+2}=xy+3$$ $$y\frac d{dx}\sqrt{x+2}+\frac{dy}{dx}\sqrt{x+2}=y\frac d{dx}x+\frac{dy}{dx}x+0$$ $$y\cdot\frac12(x+2)^{-1/2}+\frac{dy}{dx}(x+2)^{1/2}=y\cdot1+\frac{dy}{dx}x+0$$ $$\frac{dy}{dx}(x+2)^{1/2}-\frac{dy}{dx}x=-y\cdot\frac12(x+2)^{-1/2}+y$$ $$\frac{dy}{dx}\left((x+2)^{1/2}-x\right)=-\frac12y(x+2)^{-1/2}+y$$ $$\frac{dy}{dx}=\frac{-\frac12y(x+2)^{-1/2}+y}{(x+2)^{1/2}-x}$$

Please be kind; it is my first time posting here and I'm still in highschool. I'm just having trouble figuring out where I'm going wrong here.

Answer 1

There was no mistake in your work, and your final "$\frac{dy}{dx}=\frac{-\frac12y(x+2)^{-1/2}+y}{(x+2)^{1/2}-x}$" was very close to the "given answer $-\frac{y(2(x+2)^{1/2}-1)}{2(x(x+2)^{1/2}-x-2)}$" you are reporting: $$\begin{align}\frac{dy}{dx}&=\frac{-\frac12y(x+2)^{-1/2}+y}{(x+2)^{1/2}-x}\\ &=-\frac{\left(y-\frac12y(x+2)^{-1/2}\right)2(x+2)^{1/2}}{\left(x-(x+2)^{1/2}\right)2(x+2)^{1/2}}\\&=-\frac{y\left(2(x+2)^{1/2}-1\right)}{2\left(x(x+2)^{1/2}-x-2\right)}. \end{align}$$

Answer 2

Given $$y\sqrt{x+2} - xy = 3$$

Let us factor out a $y$ on the left hand side, then divide by the parenthetical term:

$$y = \frac{3}{\sqrt{x+2} - x}$$

Now, we are prepared to take the derivative with respect to $x$. This can be done e.g. using the quotient rule or product rule, along with the chain rule.

As to implicit differentiation, you could start by taking the derivative of both sides of your given equation:

$$y \cdot \frac{d}{dx}\sqrt{x+2} + \frac{dy}{dx} \cdot \sqrt{x+2} = y \cdot \frac{d}{dx}x + \frac{dy}{dx} x + 0$$

I did not write out any of the actual differentiation, since it is not clear where your troubles lie. For example, in the preceding equation, the term $\frac{d}{dx}x$ would simply be $1$. Without further context, it's unclear where you're running into trouble.

If you pursue the latter equation, you should eventually be able to rearrange so as to solve for $\frac{dy}{dx}$ in terms of $x$ alone; if done correctly, you will know because it will match up with the derivative found using the first method (standard differentiation).