Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking for a good visual way to think about partial derivatives (and slopes and tangent lines of partial derivatives) since this concept is very new for me and a little counter intuitive.

SO what is a good visual way to think about partial derivatives?

Thank you.

share|improve this question
    
khanacademy.org/math/calculus/partial_derivatives_topic It's a good & simple explanation of partial derivatives. He also shows pictures so it's pretty cool. –  Shahar Dec 15 '13 at 16:22
    
They are special cases of directional derivative. –  Mhenni Benghorbal Dec 16 '13 at 8:26

5 Answers 5

Let $f(x,y)$ be some function. Then you can visualize $\partial f / \partial x$ evaluated at a given $y=y_0$ as the derivative of a new function $g(x) = f(x, y_0)$. In particular, $\partial f / \partial x \rvert_{y=y_0}=dg/dx$.

Geometrically, visualize the graph of $f(x,y)$: some surface. Take a plane that is parallel to the $x$ and $z$ axes and slice the graph with it at a given $y=y_0$. The surface touches the plane in a curve. That curve is your $g(x)$, and taking the partial by $x$ means taking the derivative of that $g(x)$ by $x$.

Credit to alamo.edu

Taking the partial derivatives in different directions is like rotating the plane and looking at the curve in which the surface and the plane intersect.

Credit to alamo.edu for the picture.

share|improve this answer

The best way to see it is first think about fuctions $f: \mathbb{R}^2 \to \mathbb{R}$ since you can visualize this especially with computers.

You can graph this function in three dimensions and it will look like some smooth surface. Right now we can't make sense of derivatives in the usual way in one dimension. But we will be able after some restrictions. You can restrict the domain to a curve in the domain plane, say a parametric curve $\gamma: \mathbb{R} \to \mathbb{R}^2$ (some curve on the $xy$-plane). Then the resulting graph will look like a curve in three dimensions. But since you are going along a curve in the domain, the function is just $f \circ \gamma: \mathbb{R} \to \mathbb{R}$ which is a single variable function. Now you can consider the derivative of this function. This corresponds to the rate of change of the height as you go along the curve in three dimensions (actually it depends on how fast you go along $\gamma$ but we can always normalize it so that the speed is $1$, i.e. $|\gamma'| = 1$).

This is a general case. The simplest curve in the domain plane is when $\gamma$ is a straight line. Partial derivatives are the most special case. It is when $\gamma$ is not only a straight line but parallel to one of the axis ($x$-axis or $y$-axis). So its just the rate of change of the height as you go specifically in the $x$ or $y$ direction. We can't visualize this in higher dimensions but it is the same concept but with more variables.

share|improve this answer

The partial derivatices can also be seen as special cases of directional derivatives. The definition of a directional derivative in the direction $v$ is given by: $$ D_vf(x) = \lim_{h\rightarrow 0} \frac{f(x+hv)-f(x)}{h} $$ If we now look at the partial derivatives with $f$ a function of two variables then $$\frac{\partial f}{\partial x}=D_1f(x) = D_{(1,0)}f(x) = \lim_{h\rightarrow 0} \frac{f(x+(h,0))-f(x)}{h}.$$ In case of the derivative in the $y$-direction we almost get the same: $$ \frac{\partial f}{\partial y}=D_2f(x) = D_{(0,1)}f(x) = \lim_{h\rightarrow 0} \frac{f(x+(0,h))-f(x)}{h}. $$

So by studying this definitions we see that the partial derivative is the slope of one of the directions of the axis.

share|improve this answer

Consider you have a function $F(x,y,z)$. You can consider that the partial derivative of $F$ with respect to $y$ is the derivative of $F$ with respect to $y$, all other variables being fixed.

For example $F = y^2 + x y$; then, the partial derivative of $F$ with respect to $x$ is $y$ and the partial derivative of $F$ with respect to $y$ is $(2 y + x)$. we see that the partial derivative is the slope of one of the directions of the axis.

Have a look at http://en.wikipedia.org/wiki/Partial_derivative. It is pretty good.

share|improve this answer

They describe the variation of a function of multiple variables along certain directions. These directions are determined by the coordinate axis, by the very definition of the partial derivative itself. Choosing the $i$-th coordinate is equivalent to consider the $i$-th coordinate axis: as main effect, computing the partial derivative of a given function along a specific variable one keeps all other variables as "fixed".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.