# Implicit differentiation with partial derivatives?!

I really don't have idea if I am correct, but I would like to check if what is making sense to me, is indeed correct.

If I have a surface in $R^3$ that corresponds to the equation below: $$z = 6 - x - x² - 2y²$$ When I'm asked to implicit differentiate that, for example, in terms of $\frac{\partial z}{\partial y}$, why should I treat $x$ as a constant?!?!

The reason for why I should do that, in my head, is because we just want to check what a tiny change in $z$ is going to change our $y$, so it's just like taking a derivative of a curve in a plane that intersects my surface, and that plane is described by a fixed $x$ value.

Because we're fixing $x$ to get that intersection, we treat it as a constant?! Am I right?

Since: $$\frac{\partial z}{\partial y} = -4y$$ Can assume that $-4y$ is going to be the slope of the tangent line for any point of that intersection?!

For example, the point $\alpha = (0,1,4)$ is on the plane $x = 0$, and satisfies the equation for the surface. Now, if want to get the tangent line on that point, I just need to get another point that is going to follow the slope of $-4$ because $-4y : y = 1 \Rightarrow -4$, and is on the plane. That point can be, for example $\gamma = (0,2,0)$. Now I can get a director vetor $\vec{t}$, and get a parametric equation for that tangent line: $$\vec{t} = \alpha - \gamma = (0,-1,4)$$

$$x = 0 + \beta \cdot 0\\ y = 1 + \beta \cdot -1\\ z = 4 + \beta \cdot 4\\$$

Is everything correct? If it isn't, please try to explain what have I done wrong. Thanks!

• What graphing software are you using? Sep 23, 2016 at 19:37
• I'm using geogebra! why? Sep 23, 2016 at 19:40
• just to see if there's something better than Geogebra I should know about :P Sep 23, 2016 at 19:41
• Geogebra is sick hahaha! Sep 23, 2016 at 19:42

$$\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0}\ \frac{f(a+h,b) - f(a,b)}{h}$$
Therefore if you wish to compute $z_x$ then $z = z(x,y)$ and so;
$$\frac{\partial z}{\partial x}(a,b) = \frac{d}{dx} z(x,b) \Bigr|_{x=a}$$
Hence; the partial derivative in the direction of $y$ at $(a,b)$ is the same as differentiating $z(x,b)$ i.e holding $y$ as a constant.