# Intro - Partials? Now, you've mastered the art of single-variable calculus. You've climbed every hill, identified every goddamn inverse trig identity, triply-integrated by parts, and now, scarred yet roaring for more, you look to the realm of the *multivariables!* A function $f(x,y)$ is shoved before you. Two variables in, one value out. The task? To differentiate it. But how? Implicit ain't gonna work on this function - the output isn't one of our variables! So, yeah. It's the title of the article - we're gonna *partially differentiate it!* And by partially differentiating it, I mean we do so twice - one with respect to $x$, and the other with respect to $y$. We'll treat the non-differentiated variable as a constant for both - then eventually, we'll get to use them for nefarious purposes! ![[problemsetspartials.png]] *The Problem Sets. The Problem Sets.* # Fundamental Theorem of Calculus - Multivariable Edition We need to write two fundamental theorems - one for the derivative with respect to $x$, the second for the one with respect to $y$: $\frac{\delta f}{\delta x } (a,b) = \lim_{ h \to 0 } \frac{f(a+h,b) - f(a,b)}{h} \tag{1}$ $\frac{\delta f}{\delta y } (a,b) = \lim_{ h \to 0 } \frac{f(a,b+h) - f(a,b)}{h} \tag{2}$ Notice where the small increment $h$ goes for each one of these! Adding on to these, we've also got *multivariable double derivatives!* Say, if we still used $f(x,y)$ for our purposes, the different double derivatives we could write would be: $\frac{\delta^2f}{\delta x^2} = \frac{\delta}{\delta x} (\frac{\delta f}{\delta x}) \tag{3}$ $\frac{\delta^2f}{\delta y^2} = \frac{\delta}{\delta y} \frac{\delta f}{\delta y} \tag{4}$ $\frac{\delta^2f}{\delta y \delta x} = \frac{\delta}{\delta y} \frac{\delta f}{\delta x} \tag{5}$ $\frac{\delta^2f}{\delta x \delta y} = \frac{\delta}{\delta x} \frac{\delta f}{\delta y} \tag{6}$ And yes, (5) and (6) are equal to each other. You can try putting in some basic functions to see that for yourself! >[!Example]- Example Problem (MecMath Handout) >Let $u$ and $v$ be twice-differentiable functions of a single variable, and let $c$ be a constant. Show that $f (x, y)= u(x+cy)+v(x−cy)$ is a solution of the general one-dimensional wave equation: >$\frac{\delta^2f}{\delta x^2} - \frac{1}{c^2} \frac{\delta^2f}{\delta y^2} = 0 \tag{7}$ >**A:** >Easy! Differentiate to both $\frac{\delta^2f}{\delta x^2}$ and $\frac{\delta^2f}{\delta y^2}$: >$\frac{\delta^2f}{\delta x^2} = u''(x+cy) + v''(x-cy) \tag{8}$ >$\frac{\delta^2f}{\delta y^2} = c^2 u'' (x+cy) + c^2 v'' (x-cy) \tag{9}$ >...which is then trivial. Yay. # Limits of a Multivariable Function %%would rather write this later%% TBA # The Nefarious Purpose - Gradient