Implicit Differentation

Question

Did I tackle this implicit differentiation correctly?

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

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

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

Answer 1

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.

Answer 2

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.