Question

if f(x,y) is differentiable at (a,b) then it has partials at (a,b)
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Answer 1

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Since we have:

f(x,y)=x+3y−−−−−√

we take:

fx(x,y)fy(x,y)=12x+3y−−−−−√=32x+3y−−−−−√

which are both continuous - and, of course, well-defined - for every (x,y)∈R2 such that x+3y>0. So, since (1,2) is such a point then abovementioned the theorem gives us that f is differentiable at (1,2).

As for your comment about the existence of partial derivatives in an area around (a,b):

Consider the following function f:R2→R:

f(x,y)=⎧⎩⎨y∣∣∣x2sin1x∣∣∣0x≠0x=0

Also, let

A:={(x,y)∈R2∣∣∣x≠12kπ, k∈Z}

It is clear that for (x,y)∈A, fx(x,y) does not exist, however, it does exist in (0,y), for every y∈R. Indeed,

limx→0f(x,y)−f(0,y)x−0=limx→0y∣∣∣x2sin1x∣∣∣x==ylimx→0∣∣∣x2sin1x∣∣∣x==y⋅0==0

since

∣∣∣∣∣∣∣∣∣x2sin1x∣∣∣x∣∣∣∣∣∣≤|x|⇔−|x|≤∣∣∣x2sin1x∣∣∣x≤|x|

So, fx(0,y)=0, for every y∈R but fx does not exist in any are around (0,y) since it does not exist in A and all points (0,y) are accumulation points of A.

You can now take

g(x,y)=f(x,y)+f(y,x)

and have the same result for both gx and gy.

Answer 2

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Since we have:

f(x,y)=x+3y−−−−−√

we take:

fx(x,y)fy(x,y)=12x+3y−−−−−√=32x+3y−−−−−√

which are both continuous - and, of course, well-defined - for every (x,y)∈R2 such that x+3y>0. So, since (1,2) is such a point then abovementioned the theorem gives us that f is differentiable at (1,2).

As for your comment about the existence of partial derivatives in an area around (a,b):

Consider the following function f:R2→R:

f(x,y)=⎧⎩⎨y∣∣∣x2sin1x∣∣∣0x≠0x=0

Also, let

A:={(x,y)∈R2∣∣∣x≠12kπ, k∈Z}

It is clear that for (x,y)∈A, fx(x,y) does not exist, however, it does exist in (0,y), for every y∈R. Indeed,

limx→0f(x,y)−f(0,y)x−0=limx→0y∣∣∣x2sin1x∣∣∣x==ylimx→0∣∣∣x2sin1x∣∣∣x==y⋅0==0

since

∣∣∣∣∣∣∣∣∣x2sin1x∣∣∣x∣∣∣∣∣∣≤|x|⇔−|x|≤∣∣∣x2sin1x∣∣∣x≤|x|

So, fx(0,y)=0, for every y∈R but fx does not exist in any are around (0,y) since it does not exist in A and all points (0,y) are accumulation points of A.

You can now take

g(x,y)=f(x,y)+f(y,x)