How do I implicit differentiate this equation?

Question

The task is to determine the equation for the tangent line at the point $(-2,1)$ to the curve below. I would like to know if I am solving this the right way and also if the answer is right (I don't know how to solve implicit differentiates in Wolfram Alpha).

$$ x\ln y-\frac{y^2}{x+y} =1 $$

$$ \frac{d}{dx} x\ln y-\frac{y^2}{x+y} = \\ = x\frac{1}{y}y'+\ln y - y^2(-1)(x+y)^{-2}(1+y')+2y(x+y)^{-1} = \\ = x\frac{1}{y}y'+\ln y - \frac{-y^2}{(x+y)^2}(1+y')+\frac{2y}{x+y} $$

Now, I just have to set $x = -2$ and $y = 1$ and solve the equation. But I don't think that I have done the differentiation right...

Answer 1

First you need to know how to differentiate an implicant function. For example, $f(x,y)=0$ is an implicant function, and we want to know the slope of the function $y(x)$, in $f(x,y)=0$, $y$ should be regarded as a function of $x$ and then $f$ is differentiated.

In youre case, we do like this:

$f(x,y)=x\ln y-\frac{y^2}{x+y}-1=0$, and we differentiate it about $x$: $$\frac{d}{dx}f(x,y)=\frac{d}{dx}[x\ln y-\frac{y^2}{x+y}-1]\\ =\frac{d}{dx}(x\ln y)-\frac{d}{dx}(\frac{y^2}{x+y})\\ =x\frac{1}{y}y'+\ln y- \frac{2yy'(x+y)-y^2(1+y')}{(x+y)^2}\\ =(\frac{x}{y}-\frac{2y}{x+y}+\frac{y^2}{(x+y)^2})y'+lny+\frac{y^2}{(x+y)^2}\\ =0$$ then you set $x=-2,y=1$, solve the equation and have $y'$ at this point.

Answer 2

Hint

Write your function as $F(x,y)=0$. Compute the total derivative : this requires the derivative of $F(x,y)$ with respect to $x$ (at constant $y$) and the derivative of $F(x,y)$ with respect to $y$ (at constant $x$). The total derivative is $0$. Now, extract $\frac{\text{dy}}{\text{dx}}$.

In your problem, [$\frac{y^2}{(x+y)^2}+\log (y)$] and [$\frac{y^2}{(x+y)^2}-\frac{2 y}{x+y}+\frac{x}{y}$] are respectively the derivatives of $F(x,y)$ with respect to $x$ and with respect to $y$.

I am sure you can take from here. If you still have problems, just post.

Answer 3

What can help sometimes is if you think of $y$ as a function of $x$ (locally this is what we are doing) so that we write:

$$x\ln y-\frac{y^2}{x+y} =1\Rightarrow x\ln([y(x)])-\frac{[y(x)]^2}{x+[y(x)]}=1.$$

I find it is easier to see the various chain rules when we write it like this.