How would I go about solving these two questions?

How would I go about solving these two questions?

I used partial derivatives for the second (i) and get it is not differentiable as f'x=/=f'x

if by differentiaable with respect to y you mean that the partial derivative exists, well then it exists.
the limit takes place in the line x=1 therefore you use the polynomial equation

1. Write down the problem.
2. Think very hard.
3. Write down the solution.

No. Differtiating the two functions will not be equal to each other. So it's not differentiable.

I have to use both equations, as to approach from either side... but it's in 2 dimension

1. Get out my thread

it doesn't say check if it is differentiable
it says if it is differentiable with respect to y
not the same thing

for i) I think the definition is that
1) the limit as x approaches 1 (with y fixed as 1 in this case) must exist
2) f(1,1) must exist
3) f(1,1) must be equal to the limit as approaches 1 (with y fixed as 1)

Given your function, this exercise is trivial. set y=1 to get 3x^2 + 5x and 8/x

Get the limit from the right to get 8, get the limit from the left to get 8 and then compute f(1,1) to also get 1.

Continuous.

Differentiable is just like continuity but with a different limit.

Git gud at definitions, m8. This problem is literally "compute these limits".

Maybe you should learn first the definitions faggot. There isn't a limit from the right or left is R^2

>There isn't a limit from the right or left is R^2
Read the question, they are not asking you if the function is continuous at the point (1,1), they are asking you if it is continuous with respect to x at (1,1).

You are not analyzing the surface but a cross section of it, set at y=1

you are either baiting or a brainlet. enjoy calc III

O fuck exposed

I mean, if the question is just that then OP is completely retarded.

Well, it would obviously make more sense to work with the surface but how else do you interpet "with respect to x".

I am Russia :^)

C'mon this is easy. Define h(x) = 3x^2 + 5x and g(x) = 8/x, continuous polynomials. Since h(1)=8 and g(1)=8, the compound function f is continuous with respect to x at (1,1).

If x=1, then f(x,y) = 3 y^2 + 5y, and this is a differentiable function of y.

>Since h(1)=8 and g(1)=8, the compound function f is continuous with respect to x at (1,1)

>Since h(1)=8 and g(1)=8

The function being defined does not mean that the limit exists.

(i)show that the limit of the function with (x,y)->(1,1) exists and then differentiate with respect x and show that the same limit exists
(ii).do the same except that you differentiate w respect to y

OP here.
you are all fucking retarded

Just because 8=8 doesn't mean it's continous, any brainlet could sub in (1,1) "hurr durr 8=8 hence it's continous, 11 points bls".

Due to the fact it's in 3 dimensions, there are infinite paths to take and showing 2 paths doesn't prove its continuous, this would only work for 2 dimension (1 variable) function.

The first thing you have to do is take the top line and then lick a dog's asshole

Reas the question again Nigger.

Differentiating wrtx and then subbing in x=1 y=1 you get 8 and -8 respectively with either functions

Kek. You may be right, cunt.

So you are saying it is continous wrtx but not differentiable wrty?

>So you are saying it is continous wrtx but not differentiable wrty?

That is not even close to what we are trying to tell you.

You are too far below the level you cando a problem like this. Maybe go back to traditional limits for a while.

The partial derivative with respect to y does not exist at (1,1). The limit form of the derivative approaches different values depending if y is greater or less than 1. In particular this means the function is not differentiable period at (1,1) but it doesnt even ask for that.

What is it asking for then?

How do I compute f (1,1) to get 1?

Is this correct?

>inb4 brainlet

>How would I go about solving these two questions?
write down the limites. show that they exist/dont exist.

rofl you brainlet. you have to calculating lefthand and righthand limites.

But that's what he did.

>But that's what he did.
notationwise he didnt

Here you go. Maybe now you can ejaculate

Polynomials are continuous, thus both h and g are continuous at 1. Thus the left hand limit of h and the right hand limit of g exist and equal h(1) and g(1) respectively. Since h(1)=g(1), the compound function is continuous.

Dude... it literally says "with respect to x" and "with respect to y" This means for part (i) you fix y=1 and show that the function, thought of as a function of x, is continuous at x=1.