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given a function $f(x,y)$, we can easily visualize the partial derivation of $f(x,y)$ with respect to $x$ or $y$.

1. The output of the function $f(x,y)$ is in the z direction. Just like the output of $f(x)$ is in the y direction.

2. $\frac{\partial }{\partial x }f(x,y)$ can be visualized by thinking that $y$ is constant and $x$ is changing. And with the change of $x$ the output is also changing. (we can think it as the change in height of the graph or simply the change in z direction). then slight change in $z$ direction divided by slight change in $x$ direction is $\frac{\partial }{\partial x }f(x,y)$.

But what about $f(x,y,z)$?

1. If I continue to follow the previous examples about visualizing the output on a different dimension (like in $f(x,y)$ the output was in z direction) then I need to think of a fourth dimension. Which in my level, is not possible instantly, and also I don't think it's necessary to solve my problem.

2. Now to think about the partial derivation process, $\frac{\partial }{\partial z }f(x,y,z)$ means $x$ and $y$ both are to be thought as constants. Now with the change of z the output changes. But as I cannot even visualize the change of outputs as mentioned in (no 1), I cannot think of any curve forming like it did in case of $f(x,y)$ (If we pick a certainly value of $y$, say $y = 1$, and move to $x = 0$ (say) to $x = 7$ (say) and pinpoint the outputs in the $z$ direction, then adding all those points will give us a curve, which can be thought of a simple two dimensional curve and the rate of change of this very curve is what we call the partial derivative of $f(x,y)$ with respect to x.)

Now my question is simply, how do I visualize taking the partial derivative of functions that include all of the $ x, y, z $ variables?

Max
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Sami
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2 Answers2

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As it is difficult to visualize the 4th dimension, we need to think of derivatives as something other than a purely geometric construct, i.e. slope of a line.

I think of derivatives as sensitivities. If I perturb one input, holding the others steady, how much do I expect my outputs to move. This is handy when thinking about economics, where functions might have a half dozen inputs. e.g. the Back-Scholes option pricing model. The inputs are asset prices, interest rates, implied volatility and time. And "the greeks" (delta, gamma, rho, vega, theta) are the sensitivities of the option price to each of the inputs.

https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model

For a more physical example

$f(x,y,z)$ might represent temperature at a given point in space. Then the partial derivatives will give you the change in temperature for a change in position in the cardinal directions. The directional derivative gives the change in temperature for a given course, and $\nabla f(x,y,z)$ the direction to travel to maximize temperature change.

Doug M
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I think it would depend on the domain of application of the function.

For instance, in physics, a vector field F defined on an open set S is called a gradient field that can be represented as: $$F=\nabla f = (\frac{df}{dx_1},\frac{df}{dx_2},\frac{df}{dx_3}....,\frac{df}{dx_n})$$ with if there f on S. Basically, you can view this as The gradient of a scalar-valued multivariable function $f(x,y,z)$, denoted $\nabla f$, packages all its partial derivatives into a vector : $[\frac{df}{dx},\frac{df}{dy},\frac{df}{dz}]$.

In a pratical way, imagine being at $(x_0,y_0,z_0)$ the vector $\nabla f(x_0,y_o,z_0)$ tells you in wich direction you should go to increase the value of f most rapidly .

This will mean $\nabla f$ is a vector-valued function.

An example would be:

$f: R^3 \rightarrow R^3 $, which literally means we're plugging in a set $(x, y, z)$ and it the function will map it to another $(x, y, z)^*$

Let set $f(x,y,z)=(x,y,z)$. with $x>0, y>0,y>0$. Here's a applet to visualise it better: https://mathinsight.org/applet/expanding_vector_field_3D

Hope this helps.

Mahamad A. Kanouté
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