Understanding partial differential notation

Question

Say i have $$z = x^2 + y$$ $$y = 2x$$ What sense does $\frac{\partial z}{\partial x}$ make? Can we talk about changing $z$ as $x$ changes but keeping $y$ constant?

Answer 1

Since $z=z(x,y) = z(x,2x)$ I would define a function $w(x) = z(x,y(x))$. By the chain rule $$\frac{dw}{dx} = \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{dy}{dx} = 2x+2.$$ The expression for $\frac{dw}{dx}$ is called total derivative of $z$ with respect to $x$, and it is denoted by $\frac{dz}{dx}$. Be careful: in this case $z$ is a function of two variables, but the notation reminds the ordinary derivative for functions of one variable. Notice that $\partial z/\partial x$ is just a bit of the total derivative.

Answer 2

$$\frac{\partial z}{\partial x}$$ indeed denotes the variation of $z$ under a variation of $x$, while other variables are held constant.

Undisputably,

$$\frac{\partial z}{\partial x}=2x\text{ and }\frac{\partial z}{\partial y}=1.$$

If $y$ varies at the same time as $x$, you should consider a total variation

$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$

and $$\frac{dz}{dx}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\frac{dy}{dx}=2x+1\cdot2.$$

Answer 3

Yes, that’s correct. $$z = x^2+y$$ For the partial derivative $\frac{\partial z}{\partial x}$, we look at the change in $z$ while $x$ changes, while treating all other variables (in this case, $y$) as constants. $$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} x^2 + 0$$ $$\boxed{\frac{\partial z}{\partial x} = 2x}$$ Had the question been for $\frac{\partial z}{\partial y}$, we would’ve treated $x$ like a constant. $$\frac{\partial z}{\partial y} = 0+\frac{\partial}{\partial y}y$$ $$\boxed{\frac{\partial z}{\partial y} = 1}$$

Answer 4

If I understand the question, you are looking at the function $$ f(x,y) = x^2 + y $$ restricted to the line $y = 2x$. This is an example of a constrained partial derivative. Here is a video from MIT Courseware about these.

These come up in various settings, particularly in applications of calculus to physics where a function may depend on several variables that are themselves linked by physical laws. These laws may prevent us from keeping all but one variable in an expression constant. For example, if we have the constraint $PV = kT$ where $k$ is constant, and we look at $f(P,V,T)$, we cannot keep $V$ and $T$ constant while changing $P$ without violating the constraint. We have to let one of $V$ or $T$ vary, while fixing the other one.

In cases like this, we need to distinguish between the formal partial derivatives that ignore the constraint and the actual (geometrical) partial derivatives that we would actually measure using paths that stay in the constraint.

By definition, formal partial derivatives ignore the constraints, giving $$ \frac{\partial f}{\partial x} = 2x \qquad \frac{\partial f}{\partial y} = 1 $$

These can be somewhat misleading, however, because given the constraint $y = 2x$ it is impossible to fix $y$ and change $x$. The actual (geometrical) partial derivative will be measurable as the derivative along a particular curve contained in the constraint, so that $x$ changes with rate $1$ as we move along the curve.

With the constraint $y = 2x$, we are really in a one-dimensional setting. We can choose which of $x$ or $y$ will be our one independent variable. If we choose $x$, then we have $f(x) = x^2 + 2x$, and $\partial f/\partial x = 2x + 2$ is just the regular derivative.

A less degenerate example is $h(x,y,z) = 2x + 3y + z$ on the surface $x^2 + y^2 + z^2 = 1$. Now, to find $\partial h /\partial x$, we can again choose one of $y$ and $z$ to fix, because the surface is two-dimensional rather than three-dimensional. We cannot fix both $y$ and $z$ and change $x$, because $\{x,y,z\}$ is not an independent set of coordinates on the constraint surface. But we can choose which of the two ($y$ or $z$) will be fixed. Once we do that, we can compute the constrained partial derivative using the chain rule: $$ 0 = 2x\,dx + 2y\,dy + 2z\,dz\\ -2x = 2y(\partial y/ \partial x) + 2z (\partial z/\partial x) $$

In this case, there will be two different $\partial/\partial x$ derivatives, corresponding to the two choices of an independent coordinate system.

• If we let $\partial y/\partial x = 0$ then $\partial z /\partial x = -x/z$. This corresponds to following a path on the sphere parallel to the $xz$ plane. This is sometimes written $\left ( \frac{\partial z}{\partial x}\right)_y = -\frac{x}{z}$ where the $\left (\frac{\partial z}{\partial x}\right)_y$ indicates that $\{x,y\}$ is our set of independent coordinates.

• If we let $\partial z/\partial x = 0$ then $\partial y/\partial x = -x/y$. This corresponds to following a path on the sphere parallel to the $xy$ plane.

The video I linked will have more examples of this.