I need to find the implicit derivative of $y'$ with respect to x. Note that y is a function of x. I'm given: $$x^2+xy-y^2=4$$

The final result should be: $$y'=\frac{2x+y}{2y-x}$$

Here are my steps:

  1. Derivative of first term $$\frac{d}{dx}x^2 = 2x$$
  2. Derivate of second term is the product rule $$\frac{d}{dx}(xy)=x(1)y'+(1)(y)=xy'+y$$
  3. Derivative of third term $$\frac{d}{dx}-y^2=-2yy'$$
  4. Derivative of right hand side, 4, is 0.
  5. Our expression is: $$2x+xy'+y-2yy'=0$$
  6. Rewrite in terms of y'. $$xy'+y-2yy'=-2x$$ $$xy'-2yy'=-2x-y$$ $$xy'y'=\frac{-2x-y}{-2y}$$ $$y'y'=\frac{-2x-y}{-2yx}$$ $$y'^2=\frac{-2x-y}{-2yx}$$ $$y'^2=\frac{x-y}{xy}$$ $$y'=\sqrt{\frac{x-y}{xy}}$$

Have I made a mistake? I'm not sure where to go from here. I need to achieve this expression as stated at the top my question: $$y'=\frac{2x+y}{2y-x}$$

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    $\begingroup$ in step 6. I don't understand the third equality. $\endgroup$ – Jack Dec 22 '16 at 23:19
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    $\begingroup$ in step six you should collect $y'$ and then divide through $\endgroup$ – Euler_Salter Dec 22 '16 at 23:20

You've made an error in step two. Notice that


In step $6$ you've made another error. Notice that


It appears you've "divided" both sides by $-2y$, but you didn't do this properly on the left side.

If we try again:


As was desired.


Let's look at these two steps: $$xy'-2yy'=-2x-y$$ $$xy'y'=\frac{-2x-y}{-2y}$$ This is clearly not how division works. In order to divide something with more than two terms, you have to factor what you want to divide out of the expression and then divide. However, you did not factor out a $-2y$ from the expression and can not do so because not both terms have a $-2y$ in them, so this is the wrong direction to go in. Instead, try factoring out $y'$: $$y'(x-2y)=-2x-y$$ Then, to solve for $y'$, divide by $x-2y$: $$y'=\frac{-2x-y}{x-2y}$$ To simplify, multiply both the numerator and denominator by $-1$: $$y'=\frac{2x+y}{2y-x}$$


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