## Chain Rule for Partial Derivatives

1. Problem. Consider a function ${f(x,y)}$ where we parametrize
(1)$\displaystyle x=x(t,u),\quad\mbox{and}\quad y=y(t,u).$
If ${t\rightarrow t+\Delta t}$, how does ${f\rightarrow f+\Delta f}$ change? We will need to use partial derivatives and Taylor expansion

2. We first note
(2)$\displaystyle f(x+\Delta x,y) = f(x,y)+\Delta x\partial_{x}f(x,y)+\mathcal{O}(\Delta x^{2}).$
Similarly
(3)$\displaystyle f(x,y+\Delta y)= f(x,y)+\Delta y\partial_{y}f(x,y)+\mathcal{O}(\Delta y^{2}).$
Thus we find
(4)\displaystyle \begin{aligned} f(x+\Delta x,y+\Delta y) &=f(x,y+\Delta y)+\Delta x\partial_{x}f(x,y + \Delta y)+\mathcal{O}(\Delta x^{2})\\ &=\bigl(f(x,y) + \Delta y\partial_{y}f(x,y) + \mathcal{O}(\Delta y^{2})\bigr)\\ &\quad+ \Delta x\partial_{x}\bigl(f(x,y) + \Delta y\partial_{y}f(x,y) + \mathcal{O}(\Delta y^{2})\bigr)\\ &\quad+\mathcal{O}(\Delta x^{2})\\ &= f(x,y) + \Delta y\partial_{y}f(x,y) + \Delta x\partial_{x}f(x,y)\\ &\quad + \mathcal{O}(\Delta x\Delta y) +\mathcal{O}(\Delta x^{2})+\mathcal{O}(\Delta y^{2}). \end{aligned}
But specifically, we are interested in
(5)$\displaystyle \Delta x = \Delta t\partial_{t}x(t,u)+\mathcal{O}(\Delta t^{2})$
and
(6)$\displaystyle \Delta y = \Delta t\partial_{t}y(t,u)+\mathcal{O}(\Delta t^{2})$
Plugging this in allows us to write
(7)$\displaystyle \Delta f=\Delta y\partial_{y}f(x,y) + \Delta x\partial_{x}f(x,y) + \mathcal{O}(\Delta x\Delta y) +\mathcal{O}(\Delta x^{2})+\mathcal{O}(\Delta y^{2})$
as
(8)$\displaystyle \Delta f = \bigl(\Delta t\partial_{t}y\bigr)\bigl(\partial_{y}f\bigr) + \bigl(\Delta t\partial_{t}x\bigr)\bigl(\partial_{x}f\bigr) + \mathcal{O}(\Delta t^{2})$
Observe under our substitution, we have the ${\mathcal{O}(\Delta x\Delta y)}$ and other big O terms be gathered into the ${\mathcal{O}(\Delta t^{2})}$ term.

So what? Observe
(9)\displaystyle \begin{aligned} \frac{\partial f}{\partial t} &= \lim_{\Delta t\rightarrow 0}\frac{f\bigl(x(t+\Delta t,u),y(t+\Delta t)\bigr)-f\bigl(x(t,u),y(t,u)\bigr)}{\Delta t} \\ &=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}. \end{aligned}
This is precisely the chain rule. Similarly, we find
(10)$\displaystyle \frac{\partial f}{\partial u} =\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}.$

3. Implicit Differentiation Revisited. Recall implicit differentiation required us to find ${\mathrm{d} y/\mathrm{d} x}$ from some complicated expression like
(11)$\displaystyle \mathrm{e}^{xy}+4y^{2}+\tan(x+y)=0$
What to do? First we write
(12)$\displaystyle z = F(x,y) = \mathrm{e}^{xy}+4y^{2}+\tan(x+y)=0.$
Next we say ${y=y(x)}$. So we find
(13)\displaystyle \begin{aligned} \frac{\mathrm{d} z}{\mathrm{d} x} &= \frac{\partial F(x,y)}{\partial x}\frac{\mathrm{d} x}{\mathrm{d} x}+\frac{\partial F(x,y)}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &=0 \end{aligned}
where we set the derivative of ${F}$ to be zero since it’s equal to the derivative of zero. We can then write (taking ${\mathrm{d} x/\mathrm{d} x=1}$)
(14)$\displaystyle -\frac{\partial F(x,y)}{\partial x}=\frac{\partial F(x,y)}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x}$
and divide both sides by ${\partial_{y}F}$ to get
(15)$\displaystyle \frac{-\partial_{x}F}{\partial_{y}F} = \frac{\mathrm{d} y}{\mathrm{d} x}.$
But this is precisely what implicit differentiation gives us!

4. Warning for Physicists. Physicists often use partial derivative notation slightly differently. If ${q(t)}$ is the position of a particle, and ${p(t)}$ is its momentum, physicists consider arbitrary functions of the form
(16)$\displaystyle f=f(q,p,t)$
and write
(17)$\displaystyle \frac{\partial f}{\partial t} = \lim_{\Delta t\rightarrow0} \frac{f\bigl(q(t),p(t),t+\Delta t\bigr)-f\bigl(q(t),p(t),t\bigr)}{\Delta t}.$
This is strictly speaking not quite true. The error committed lies in treating ${q}$ and ${p}$ as functions of time: really they are variables whom we are trying to express as functions of time.

Exercise 1. Let ${w=\sqrt{x}+y^{2}/z}$ where ${x=\exp(2t)}$, ${y=t^{3}+4t}$, and ${z=t^{2}-t}$. Find ${\mathrm{d} w/\mathrm{d} t}$.

Exercise 2. Let ${z=\cos(xy)+y\sin(x)}$ where ${x=v^{2}+u}$ and ${y=u-v}$. Find ${\partial_{u} z}$ and ${\partial_{v}z}$.

Exercise 3 [Math.SE]. Consider the equation ${\displaystyle f(x,y)=\cos(x^2)+2xy+\sin(y^2)-4x-1+y=0}$. What is ${\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}$?