# Implicit Differentation

Did I tackle this implicit differentiation correctly?

$$5x^2+3xy-y^2=5$$

$$10x+3x\dfrac{dy}{dx}+3y-2y\dfrac{dy}{dx}=0$$ $$\dfrac{dy}{dx}(y-2y)=-10x-3z-3y$$

$$\dfrac{dy}{dx}=\dfrac{-10x-3x-3y}{y-2y}$$

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A couple of algebra slips. –  André Nicolas Nov 22 '12 at 20:00

Unfortunately not. There seems to be a mistake in your second line. Let me take you through the steps:

$$5x^2+3xy-y^2 = 5 \, ,$$ $$10x + 3y + 3x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0 \, ,$$ $$(3x-2y)\frac{dy}{dx} = -10x - 3y \, ,$$ $$(2y-3x)\frac{dy}{dx} = 10x + 3y \, ,$$ $$\frac{dy}{dx} = \frac{10x+3y}{2y-3x} \, .$$

Of course, this is not satisfactory. There should not be any $y$s on the right-hand-side. From the definition, we can obtain an expression for $y$:

$$y = \frac{3}{2}x \pm \frac{1}{2}\sqrt{29x^2-20} \, .$$

Putting this into the formula for $dy/dx$, we get:

$$\frac{dy}{dx} = \frac{\pm29x+3\sqrt{29x^2-20}}{\sqrt{29x^2-20}}$$

This is only well defined for $29x^2-20 > 0.$ When $29x^2-20=0$ the hyperbola will have vertical tangents, and so $dy/dx$ is not well-defined. When $29x^2-20 < 0$, there will be no corresponding $y$, i.e. you will be "inside" the hyperbola.

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Why did −10x become 10x? –  soniccool Nov 22 '12 at 19:53
I multiplied both sides by $-1$, so $3x-2y$ became $-3x + 2y = 2y - 3x$ and $-10x-3y$ became $10x+3y$. You would still get the same answer, I just like to get rid of minus signs if possible. –  Fly by Night Nov 22 '12 at 19:56
$\frac{dy}{dx}$ above which you got is completely satisfactory, because the OP wants to take an implicit diff. Look at the tag the OP added. Of course what you were concluding about existing a function like $y$ of variable $x$ is completely right. I think, the OP has been sure of existing a function like $y$. Any way, when $y$, a function of $x$, both are included in a relation, then what you did step by step is formal and is right. –  Babak S. Nov 22 '12 at 20:11

Beside to @Andre's answer, if you are familiar to partial differentiation then you can use the following rule: $$y'=-\frac{F_x(x,y)}{F_y(x,y)}$$ where in we assume our relation can be written as $F(x,y)=0$. Note that $y$ should be a function of $x$ at this rule.

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$+1 \quad +^{\Large\ddot\smile}$ –  amWhy Apr 6 '13 at 3:17