# Real-valued function defined by the Cartesian basis vectors…

If ${e_x}$ and ${e_y}$ denote the unit Cartesian basis vectors and for a real-valued function $f(x,y)$, $$\nabla f = ({{\delta f} \over {\delta x}},{{\delta f} \over {\delta y}}) \equiv {{\delta f} \over {\delta x}}{e_x} + {{\delta f} \over {\delta y}}{e_y}$$ In polar coords, $x = r\cos \theta$ & $y = r\sin \theta.$

We have $${e_r} = \cos \theta {e_x} + \sin \theta {e_y}$$ $${e_\theta } = - \sin \theta {e_x} + \cos \theta {e_y}$$ $$F(r,\theta ) = f(x(r,\theta ),y(r,\theta ))$$ Using the chain rule, show $${{\delta F} \over {\delta r}} = \cos \theta {{\delta f} \over {\delta x}} + \sin \theta {{\delta f} \over {\delta y}}$$ Therefore show that $\nabla f = {{\delta F} \over {\delta r}}{e_r} + {1 \over r}{{\delta F} \over {\delta \theta }}{e_\theta }$...

I don't know how to get that last step.

$r = \sqrt{x^2+y^2}$ $\frac{\partial r}{\partial x} =$$\frac{x}{\sqrt{x^2+y^2}}\\ \frac{x}{r}\\ \frac{r\cos\theta}{r}\\ \cos\theta \frac{\partial r}{\partial y} = \sin\theta \theta =\tan^{-1}(\frac{y}{x}) \frac{\partial \theta}{\partial x} =$$ \frac{-y}{x^2+y^2}\\ \frac{-\sin \theta}{r}$

$\frac{\partial \theta}{\partial y} =$$\frac{x}{x^2+y^2}\\ \frac{cos \theta}{r} \nabla f =$$ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x},\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y}\\ \frac{\partial f}{\partial r}\cos\theta - \frac{\partial f}{\partial \theta}\frac{\sin\theta}{r},\frac{\partial f}{\partial r}\sin\theta + \frac{\partial f}{\partial \theta}\frac{\cos\theta}{r}\\ \frac{\partial f}{\partial r} e_r+\frac{\partial f}{\partial \theta} \frac{e_\theta}{r}$

Doug M's way is better/neater, but here's an alternative:

$$\Delta f = \frac{\partial F}{\partial r}e_r + \frac{1}{r}\frac{\partial F}{\partial \theta}e_\theta$$

From your previous result, another application of the chain rule for $\partial F / \partial \theta$ and expanding:

$$\Delta f = \cos^{2}(\theta)\frac{\partial f}{\partial x}e_x + \sin(\theta)\cos(\theta)\frac{\partial f}{\partial x}e_y + \sin(\theta)\cos(\theta)\frac{\partial f}{\partial y}e_x + \sin^{2}(\theta)\frac{\partial f}{\partial y}e_y + \sin^{2}(\theta)\frac{\partial f}{\partial x}e_x - \sin(\theta)\cos(\theta)\frac{\partial f}{\partial y}e_x - \sin(\theta)\cos(\theta)\frac{\partial f}{\partial x}e_y + \cos^{2}(\theta)\frac{\partial f}{\partial y}e_y$$

Cancelling terms and using $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$ you get:

$$\Delta f = \frac{\partial f}{\partial x}e_x + \frac{\partial f}{\partial y}e_y$$