# MIT OCW

## 18.02 Multivariable Calculus

### Total Differentials

When we are considering a multivariable function, is there a way to hold changes of all components?

Well there's the total differential defined for `$f(x,y,z)$`

that is `$$\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z.$$`

**Notice** I've been confusing the derivatives and differentials, but I'm now clearing the edge between them. In single variable situation, when apply "differential" to some function `$f(x)$`

, we get actually another function `$$\mathrm{d}f(x,\Delta x)\overset{\Delta}{=}f'(x)\Delta x.$$`

We often write something like `$$\mathrm{d}f=\boxed{}\mathrm{d}x$$`

because according to definition `$$\mathrm{d}(x,\Delta x)=\Delta x.$$`

### Chain Rule

If we have some multivariable function `$f(x,y,z)$`

, where `$x=x(t),y=y(t),z=z(t)$`

, we can get `$$\frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{\partial f}{\partial z}\frac{\mathrm{d}z}{\mathrm{d}t}.$$`

#### Validation for Product and Quotient Rule

Treat product of two functions `$u=u(t),v=v(t)$`

as a multivariable function `$f(u,v)=uv$`

, and apply the chain rule `$$\begin{aligned}\frac{\mathrm{d}f}{\mathrm{d}t}&=\frac{\partial f}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial f}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t}\\&=v\frac{\mathrm{d}u}{\mathrm{d}t}+u\frac{\mathrm{d}v}{\mathrm{d}t}.\end{aligned}$$`

The quotient rule can be validated similarly, omitted.

### Chain Rule for Several Variables

Given a function `$f(x,y)$`

where `$x=x(u,v),y=y(u,v)$`

, how to get `$\frac{\partial f}{\partial u}$`

and `$\frac{\partial f}{\partial v}$`

without plugging `$x=x(u,v)$`

and `$y=y(u,v)$`

in?

Let's calculate the total differential of `$f$`

, that is `$$\begin{aligned}\mathrm{d}f&=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y\\&=\frac{\partial f}{\partial x}\left(\frac{\partial x}{\partial u}\mathrm{d}u+\frac{\partial x}{\partial v}\mathrm{d}v\right)+\frac{\partial f}{\partial y}\left(\frac{\partial y}{\partial u}\mathrm{d}u+\frac{\partial y}{\partial v}\mathrm{d}v\right)\\&=\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\right)\mathrm{d}u+\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}\right)\mathrm{d}v.\end{aligned}$$`

And notice that `$$\mathrm{d}f=\frac{\partial f}{\partial u}\mathrm{d}u+\frac{\partial f}{\partial v}\mathrm{d}v,$$`

therefore we get `$$\left\{\begin{aligned}\frac{\partial f}{\partial u}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\\\frac{\partial f}{\partial v}&=\frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}.\end{aligned}\right.$$`