Solving problems using the *definition* of differentiability There is a problem in my textbook, that I could not solve and was not able to understand the solution to. The problem had part a, b, c, d. Only a were solved. I am out of luck. I hope, if somebody writes the solution so one of these, I might understand it.
Question:
Use the definition of differentiability to prove that the following functions are in fact differentiable @ each point.


*

*b. $(1+x+2y)^2 \quad @ (1,1)$

*c. $e^{x+2y} \quad \quad \quad @ (2,2)$

*d. $\sin(x+y) \quad @ (1,1)$

 A: You should try to prove that
$$\lim_{(x,y)\rightarrow (x_0,y_0)}\frac{f(x,y)-f(x_0,y_0)-\langle\nabla f(x_0,y_0),(x-x_0,y-y_0)\rangle}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0,$$
denoting by $\nabla f(x_0,y_0)$ the gradient of $f$ at $(x_0,y_0)$, in all 3 cases.
Let us try with
$$f(x,y)=(1+x+2y)^2 $$
and $(x_0,y_0)=(1,1)$. A quick computation gives 
$$\nabla f(1,1)=(2(1+1+2),4(1+1+2))=(8,16).$$
Then our limit becomes
$$\lim_{(x,y)\rightarrow (1,1)}\frac{f(x,y)-f(1,1)-\langle\nabla f(1,1),(x-1,y-1)\rangle}{\sqrt{(x-1)^2+(y-1)^2}}=\lim_{(x,y)\rightarrow (1,1)}\frac{(1+x+2y)^2-16-8(x-1)-16(y-1)}{\sqrt{(x-1)^2+(y-1)^2}}=\lim_{(x,y)\rightarrow (1,1)}\frac{9+x^2+4y^2-6x-12y+4xy}{\sqrt{(x-1)^2+(y-1)^2}}=\text{magic trick: I highlight} (x-1)^2+(y-1)^2 \text{in the numerator}=\lim_{(x,y)\rightarrow (1,1)}\frac{(x-1)^2+(y-1)^2+3y^2-4x-10y+7+4xy}{\sqrt{(x-1)^2+(y-1)^2}}=
\lim_{(x,y)\rightarrow (1,1)}\left[\sqrt{(x-1)^2+(y-1)^2}+\frac{3y^2-4x-10y+7+4xy}{\sqrt{(x-1)^2+(y-1)^2}}\right]=0, $$
as the second addendum in polar coordinates $(x,y)=(1+\rho\cos\theta,1+\rho\sin\theta)$ around $(1,1)$ becomes
$$ 
\frac{3(1+\rho\sin\theta)^2-4(1+\rho\cos\theta)-10(1+\rho\sin\theta)+7+4(1+\rho\cos\theta)(1+\rho\sin\theta)}{\rho}=\frac{O(\rho^2)}{\rho}.$$
The other cases are treated similarly. You must check that the gradient of $f$ at the given point exists and proceed with the computations.
A: I will solve one as you asked. The derivative of function (b) $f(x,y)=(1+x+2y)^2$ is $$\bigg(2(1+x+2y),4(1+x+2y)\bigg)$$ So the difference between the function and its linear approximation (derivative times point plus function value) is $$(1+x+2y)^2-\bigg(2(1+x+2y),4(1+x+2y)\bigg)\cdot (x-1,y-1)-f(1,1)=(1+x+2y)\bigg(1+x+2y-2(x-1)-4(y-1)\bigg)-16=(1+x+2y)(7-x-2y)-16$$ The relevant limit is therefore $$\lim_{(x,y)\to (1,1)}{\frac{(1+x+2y)(7-x-2y)-16}{\sqrt{(x-1)^2+(y-1)^2}}}$$ This equals zero, which can be shown by switching to polar coordinates (note that the degree of the numerator is 2 and the denominator only 1).
