The Student Room Group

Interview question

What do I do if this happens? I think at least my interviewer was wrong so I just did it how I thought he wanted it an carried on.

He gave me
f(x,y,z) = xy/(z^2)

then he told me to find

f'
I asked him if he meant partial differentiation but he said no? was I wrong? I just differentiated with respect with x,y and z then added them.

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Original post by isiaiah d
partial differentiation but he said no?

i dont even know what partial differentiation means?
What interview is this ?
Original post by isiaiah d
I just differentiated with respect with x,y and z then added them.

Sounds like a good approach. But I dont know if it is right ?
Reply 3
Original post by Rohan77642
Sounds like a good approach. But I dont know if it is right ?


yeah I have no idea, I've never seen a question like that so im unsure if I'm wrong or he was. I also asked him how he would write the question in the dy/dx or df/... form and he didn't really say anything I could understand so I'm assuming that means he was wrong?
Reply 4
Original post by Rohan77642
i dont even know what partial differentiation means?
What interview is this ?


Was chemical engineering at manchester so wasn't really expecting it either
Original post by isiaiah d
yeah I have no idea, I've never seen a question like that so im unsure if I'm wrong or he was. I also asked him how he would write the question in the dy/dx or df/... form and he didn't really say anything I could understand so I'm assuming that means he was wrong?

where was this? Imperial ?
Original post by isiaiah d
yeah I have no idea, I've never seen a question like that so im unsure if I'm wrong or he was. I also asked him how he would write the question in the dy/dx or df/... form and he didn't really say anything I could understand so I'm assuming that means he was wrong?


I thought dy/dx is f'
What do you mean by partial differentiatiom?
Reply 7
Original post by AzureCeleste
I thought dy/dx is f'
What do you mean by partial differentiatiom?


well he just wrote f'.
dy/dx is Leibniz notation, since he said with respect to everything I wasn't sure what the leibniz notation would look like.

What I did was just

df/dx + df/dy + df/dz
Reply 8
Original post by Rohan77642
where was this? Imperial ?


manchester, I have imperial later this month so wanted to ask what to do next time if I thought the interviewer was wrong
Original post by isiaiah d
well he just wrote f'.
dy/dx is Leibniz notation, since he said with respect to everything I wasn't sure what the leibniz notation would look like.

What I did was just

df/dx + df/dy + df/dz


I mean from my experience at school, from what I remember. f' is the exact same as dy/dx, just a different way of writing it
Original post by isiaiah d
Was chemical engineering at manchester so wasn't really expecting it either

Damn!
Idk how to do the question, but I would have done something similar to what you did I guess if that sounds reassuring. Best of luck though!
Original post by isiaiah d
manchester, I have imperial later this month so wanted to ask what to do next time if I thought the interviewer was wrong

Idk probably politely disagree or argue your point? Idk, I have no experience with interviews.
Reply 12
Original post by AzureCeleste
I mean from my experience at school, from what I remember. f' is the exact same as dy/dx, just a different way of writing it


it is the same but you would have to define the function f(x) = y or you would write f'(x) = df/dx.

What he wrote though was

f(x,y,z) = xy/(z^2)

and then told me to find
f'

which as far an I'm aware doesn't make sense?
Reply 13
Thinking about a slightly easier example: f(x, y) would be a function which can be visualized in three dimensions: x, y and f(x,y). You can think of f' as the slope of the plane, tangent to the function at (x, y) . If y is kept constant while x is varied (ie df/dy = 0), f' would be equal to df/dx, and if x is kept constant and y varies (df/dx = 0), f' would be df/dy. If both x and y are free to vary, then it makes sense from a graphical point of view for f' to be linearly related to both df/dx and and df/dy, meaning that f' = df/dx plus df/dy, satisfying the functions we previously got when df/dx = 0 and when df/dy = 0. Extrapolating this to three variables x, y and z: f' = df/dx plus df/dy plus df/dz.
Hope this somewhat helped.

Edit:
Sorry about the "plus"es, my phone doesn't seem to be able to write plus signs for some reason
(edited 5 years ago)
Reply 14
Original post by Jono137
Thinking about a slightly easier example: f(x, y) would be a function which can be visualized in three dimensions: x, y and f(x,y). You can think of f' as the slope of the plane, tangent to the function at (x, y) . If y is kept constant while x is varied (ie df/dy = 0), f' would be equal to df/dx, and if x is kept constant and y varies (df/dx = 0), f' would be df/dy. If both x and y are free to vary, then it makes sense from a graphical point of view for f' to be linearly related to both df/dx and and df/dy, meaning that f' = df/dx df/dy, satisfying the functions we previously got when df/dx = 0 and when df/dy = 0. Extrapolating this to three variables x, y and z: f' = df/dx df/dy df/dz.
Hope this somewhat helped.


so multiplying the partial derivatives together then or when you say df/dx do you mean the full derivative and I was meant to use implicit differentiation?
Reply 15
Original post by isiaiah d
so multiplying the partial derivatives together then or when you say df/dx do you mean the full derivative and I was meant to use implicit differentiation?


There's supposed to be plus signs in between them. Sorry about that, TSR on mobile isn't great for posting.

What I meant to say was that what you wrote down was correct.
(edited 5 years ago)
Reply 16
Original post by Jono137
There's supposed to be plus signs in between them. Sorry about that, TSR on mobile isn't great for posting.


ahh yeah, thats what I was thinking too, at first I thought the question doesn't make sense but maybe it's his wording because graphically t does make sense for there to be a derivative with respect for more than 1 variable, I guess I've just never come across it yet. I also don't fully understand why you would just add them. Thinking about it if you have a function with 2 variables the derivative would also have 2 variables if both are changing
Reply 17
Reading into it a bit more, it seems we're on the right lines, except that instead of straight up adding the partial derivatives up, you add up each partial derivative times its unit vector in that direction, so f' = (df/dx) i plus (df/dy) j plus (df/dz) k. Quite interesting stuff! Here's a Wikipedia link if you want to read more:
https://en.m.wikipedia.org/wiki/Gradient
I'm assuming you've done vectors before?
Reply 18
Original post by Jono137
Reading into it a bit more, it seems we're on the right lines, except that instead of straight up adding the partial derivatives up, you add up each partial derivative times its unit vector in that direction, so f' = (df/dx) i plus (df/dy) j plus (df/dz) k. Quite interesting stuff! Here's a Wikipedia link if you want to read more:
https://en.m.wikipedia.org/wiki/Gradient
I'm assuming you've done vectors before?


yeah ive done up to A2 so ill check this out tommo morning actually does sound interresting
Reply 19
Original post by isiaiah d
What do I do if this happens? I think at least my interviewer was wrong so I just did it how I thought he wanted it an carried on.

He gave me
f(x,y,z) = xy/(z^2)

then he told me to find

f'
I asked him if he meant partial differentiation but he said no? was I wrong? I just differentiated with respect with x,y and z then added them.


Just an idea of mine to throw out. they maybe expected you to find ∇f (grad of f) where f is a scalar field. notation wise doesn't look good but they have very similar meaning and if that was what they were looking then the solution would be a vector as shown:

(edited 5 years ago)

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