Using the definition of derivative, show that $f(x,y)=5x^2+7xy$ is differentiable at $(1,2)$ Here is what I got.
I need to use the definition, so I need to show 2 things
$$
f(x,y)=f(x_0,y_0)+ \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+ \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)+r(x,y)
$$
and 
$$
\lim_{(x,y)\to(1,2)} \frac{r(x,y)}{\sqrt{(x-1)^2+(y-2)^2}}=0.
$$
Now
$$
\frac{\partial f}{\partial x}=10x+7y  \quad\Rightarrow\quad\frac{\partial f}{\partial x}(1,2)=24,
$$
$$
\frac{\partial f}{\partial y}=7x   \quad\Rightarrow\quad\frac{\partial f}{\partial y}(1,2)=7.
$$
$$
f(1,2)=19
$$
I substitute back in , I get 
$$
f(x,y)=19+ 24(x-1)+ 7(y-2)+r(x,y)
$$
now I'm stuck. What should I do next to prove 
$$\lim_{(x,y)\to (1,2)} \frac{r(x,y)}{\sqrt{(x-1)^2+(y-2)^2}}=0?$$
 A: $$f(x,y)=19+ 24(x-1)+ 7(y-2)+r(x,y)\iff$$
$$r(x,y)=f(x,y)-24(x-1)-7(y-2)-19=5x^2+7xy-24(x-1)-7(y-2)-19$$
$$\implies\frac{r(x,y)}{\sqrt{(x-1)^2+(y-2)^2}}=\frac{5x^2+7xy-24(x-1)-7(y-2)-19}{\sqrt{(x-1)^2+(y-2)^2}}$$
Now you can use "moved" polar coordinates (or simply a substitution, if you will):
$$x-1=r\cos\theta\;,\;\;y-2=r\sin\theta\implies$$
$$\implies\frac{5(1+r\cos\theta)^2+7(1+r\cos\theta)(2+r\sin\theta)-24r\cos\theta-7r\sin\theta-19}r=$$
$$=\color{purple}{10\cos\theta}+5r\cos^2\theta+\color{green}{7\sin\theta}+\color{purple}{14\cos\theta}+7r\cos\theta\sin\theta-\color{purple}{24\cos\theta}-\color{green}{7\sin\theta}=$$
$$=r\left(5\cos^2\theta+7\cos\theta\sin\theta\right)\xrightarrow[r\to 0]{}0$$
A: $f(x,y)=5x^2 + 7xy$
Differenciate with respect to $x$ and $y$
With respect to x where y is a constant:
$10x + 7y$
With respect to y where x is a constant:
$7x$
You can evaluate both of these expressions at $x=1$, $y=2$
$10\times1 + 7\times2 = 24$
$7\times1 = 7$
So the matrix $[24\:\:\:\:7]$ is the differential at $(1,2)$
