What is a partial derivative?
In previous lessons, we learned how derivatives
of the form \(dy/dx\) (or \(f'(x)\)) represent the steepness of a function \(f(x)\) at each \(x\)-value. Ordinary derivatives give the steepness at each point along a curve. But suppose instead that we wanted to represent the steepness at each point along a surface, \(z=f(x,y)\). How would we do that? Unlike in single-variable calculus where we used a single kind of derivative operator to represent the slope at each point, in multi-variable calculus there are more than one different kinds of derivatives which we use to describe the steepness at each point along a surface. The first kind of derivative of multi-variable functions the we'll study are called partial derivatives. To understand the meaning of partial derivatives, let's star out by looking at the function
$$f(x,y)=x^2+y^2.\tag{1}$$
Figure 1: Graph of the surface \(f(x,y)=x^2+y^2\). Image courtesy of Wolfram Alpha.
A graph of this surface is illustrated above. As you can see from the graph above, this function forms a bowl-shaped surface. Let's set \(y\) equal to some constant, say \(y=1\), so that Equation (1) becomes
$$f(x,1)=x^2+1.\tag{2}$$
Figure 2: The entire blue surface is given by the function \(f(x,y)=x^2+y^2\). By letting \(y=1\), we that \(y^2=1\) and \(f(x,1)=x^2+1\) giving us a parabola shifted up one unit along the \(z\)-axis. That is how we can analytically obtain the parabola \(f(x,y)\). We can also obtain \(f(x,1)\) by passing the plane \(y=1\) (illustrated as the black plane above) through the surface \(f(x,y)\). The points at which the two surfaces (the surface \(f(x,y)\) and the plane \(y=1\)) intersect form the red parabola drawn in the image above. Image credit: By IkamusumeFan (Own work) [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons.
That's how you can obtain Equation (1) analytically. But, visually speaking, you can obtain the graph of \(f(x,1)\) by taking the plane \(y=1\) and "slicing" it through the function \(f(x,y)\) as illustrated in Figure 2. The portion of the surface \(f(x,y)\) that passes through the plane \(y=1\) is the parabola \(f(x,1)\). If we took the derivative of Equation (2) with respect to \(x\), we would get
$$f'(x,1)=2x.\tag{3}$$
Equation (3) tells you the steepness at each point along the parabola \(f(x,1)\) as you move along this parabola by varying \(x\). Equation (3) is the partial derivative of \(f(x,y)\) with respect to \(x\) evaluated at \(y=1\). Thus, partial derivatives aren't anything new and are no different from ordinary derivatives that you encountered in single-variable calculus and in previous lessons. To evaluate the partial derivative of any surface \(z=f(x,y)\) with respect to \(x\), we just take the ordinary derivative of \(f(x,y)\) with respect to \(x\) with \(y\) set equal to a constant. In other words, all the partial derivative of an arbitrary surface with respect to \(x\) is is just
$$\frac{dz}{dx}\biggl|_{y=constant}=f'(x, constant).\tag{4}$$
However, since the notation in Equation (4) is quite cumbersome, the typical notation used to represent the partial derivative of a surface \(z=f(x,y)\) is
$$\frac{∂z}{∂x}=\frac{∂}{∂x}f(x,y)$$
or
$$z_x=f_x(x,y).$$
Either notation would work just fine. To get the partial derivative of \(f(x,y)=x^2+y^2\) with respect to \(x\), we just take the ordinary derivative of this function while setting \(y\) equal to a constant. Since the ordinary derivative of a constant is zero, the second term becomes zero when we take the partial derivative. Taking the ordinary derivative of \(x^2\) with respect to \(x\), we just get \(2x\). Thus, the partial derivative of \(f(x,y)=x^2+y^2\) with respect to \(x\) just becomes
$$f_x(x,y)=2x.\tag{5}$$
If we wish to consider the partial derivative of \(x^2+y^2\) with respect to \(x\) evaluated at \(y=1\), then we must set the variable \(y\) in the function \(f_x(x,y)\) equal to one. But since there is no \(y\)-variable in Equation (5), the partial derivative of \(f(x,y)\) with respect to \(x\) is the same as the partial derivative of \(f(x,y)\) evaluated at \(y\). The notation that we would use to represent the partial derivative of an arbitrary surface \(z=f(x,y)\) with respect to \(x\) evaluated at \(y=1\) is
$$\frac{∂z}{∂x}\biggl|_{y=1}=\frac{∂}{∂x}f(x,y)\biggl|_{y=1}$$
or
$$z_x\biggl|_{y=1}=f_x(x,1).$$
Thus, we can rewrite Equation (3) as
$$f_x(x,1)=2x.$$
Finding the partial derivative of \(\textbf{f(x,y)}\) with respect to \(\textbf{y}\)
Let's now talk about the notion of the partial derivative of a surface \(f(x,y)\) with respect to \(y\) which is written as either
$$\frac{∂z}{∂y}=\frac{∂}{∂y}f(x,y)$$
or
$$z_y=f_y(x,y).$$
All that a partial derivative of a surface \(f(x,y)\) with respect to \(y\) is is the ordinary derivative of \(f(x,y)\) with respect to \(y\) with \(x\) set equal to some constant. In other words,
$$f_y(x,y)=f'(constant, y).$$
Let's suppose that we wanted to evaluate the partial derivative of \(f(x,y)=x^2+y^2\) with respect to \(y\). Using the formula above, we'd get
$$f_y(x,y)=\frac{d}{dx}\biggl(x^2+(constant)^2\biggr)=2y.$$
To obtain \(f_y(x,y)=2y\), all we had to do was treat the \(x^2\) term as a constant and take the ordinary derivative of the expression \(x^2+y^2\) with respect to \(y\). Notice that since \(f_y(x,y)=2y\) does not depend on \(x\), we could slice the surface \(f(x,y)\) with any plane \(x=k\) (where \(k\) is some constant) and each parabola on each plane would have the same derivative function, \(f_y(x,y)=f'(x,k)=2y\), describing how its steepness varied with the variable \(y\).
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