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}}?

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Derivative, Partial Derivative and tagged , . Bookmark the permalink.

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