Chain Rule ( minutes)

Recall: $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

Also true if other variables are used $$df=\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial v}dv$$

So, what if u, v are functions of x, y?

$$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$$

$$dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy$$

$$\Rightarrow df=\frac{\partial f}{\partial u}(\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy)+\frac{\partial f}{\partial v}(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy)$$ $$ df=(\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x})dx+(\frac{\partial f}{\partial u}\frac{\partial u}{\partial y} +\frac{\partial f}{\partial v}\frac{\partial v}{\partial y})dy$$

$$ \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$$

$$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y} +\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$$

Example1: Polar Coordinates $$x=rcos\phi, \, y=rsin\phi$$

Substitution $$r=\sqrt{x^2+y^2}$$ $$\phi={tan^{-1}\frac{y}{x}}$$ $$\frac{\partial r}{\partial x}=\frac{1}{2\sqrt{x^2+y^2}}\dot 2x$$

Differentials $$r^2=x^2+y^2$$ $$tan \phi=\frac{y}{x}$$ $$2rdr=2xdx+2ydy$$ $$sec^2\phi d\phi=(1+tan^2 \phi) d\phi=\frac{xdy-ydx}{x^2+y^2}$$ $$d\phi=\frac{1}{1+\frac{y^2}{x^2}}\frac{xdy-ydx}{x^2}=\frac{xdy-ydx}{x^2+y^2}$$ $$\frac{\partial \phi}{\partial x}=\frac{-y}{x^2+y^2}$$

Example2: $$\frac{\partial}{\partial}(xy)$$ Substitution $$xy=r^2sin\phi cos\phi$$ $$\Rightarrow \frac{\partial}{\partial \phi}(xy)=r^2(cos^2\phi-sin^2\phi)$$ Formula $$\frac{\partial(xy)}{\partial \phi}=\frac{\partial (xy)}{\partial x}\frac{\partial x}{\partial \phi}+\frac{\partial (xy)}{\partial y}\frac{\partial y}{\partial \phi}$$ $$\frac{\partial(xy)}{\partial \phi}=y(-rsin \phi)+x(rcos\phi)$$ $$\frac{\partial(xy)}{\partial \phi}=-r^2sin^2\phi+r^2cos^2\phi$$

Divide $$d(xy) = ydx +xdy$$ $$\Rightarrow \frac{\partial x}{\partial \phi} = y \frac{\partial x}{\partial \phi}+x\frac{\partial y}{\partial \phi}=y(-rsin\phi)+x(rcos\phi)$$

Differentials

$$d(xy)=ydx+xdy$$ $$ = y(drcos\phi-rsin\phi d\phi)+x(drsin\phi+rcos\phi d\phi)$$ $$ = 2rsin\phi cos\phi dr + r^2(cos^2\phi-sin^2\phi)d\phi$$ $$\Rightarrow \frac{\partial(xy)}{\partial\phi}=r^2(cos^2\phi-cos^2\phi)$$