# Introduction to Partial Derivatives

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 $$y$$.