Question

Differentiation help. *question attached below.

Answer 1

\[\frac{ \partial f}{ \partial y} =\frac{d f(y)}{dy}\]
just pretend x and z are constants went differentiating w.r.t y

Answer 2

oops i'm looking at b

Answer 3

i should be looking at a?

Answer 4

\[grad f=\frac{ \partial f}{ \partial x}i+\frac{\partial f}{\partial y}j+\frac{ \partial f}{\partial z} k\]

Answer 5

to find the partial of f w.r.t x pretend like y and z are constants for now

Answer 6

oh and there is no z up there so we just need the first two addends there

Answer 7

the first question is weird because
ln(0) doesn't exist...

Answer 8

@iTiaax are you there?

Answer 9

Sorry, I was working on another question

Answer 10

first question doesn't make sense to me
because when I check to see if the point is on that relation
it isn't

Answer 11

Um, why are we applying partial derivatives?

Answer 12

because I got confused because i started looking at part b instead

Answer 13

lol

Answer 14

I know the answer for that is undefined, but I was having trouble with differentiating it before making my substitutions

Answer 15

so i was like well we could do that here but we don't have to

Answer 16

the problem @iTiaax I'm having is (1,0) isn't on that relation

Answer 17

ln(0) doesn't exist
so that tells me I don't even need to differentiate it

Answer 18

\[\ln(1 \cdot 0^2)-\sin(0)=3(1)-2(0) \\ \ln(0)-0=3 \\ \ln(0)=3 \\ \text{ \not true }\]

Answer 19

thought if you do want to find y' ... anyhow we can do that

Answer 20

but there is no point

Answer 21

\[\frac{d}{dx} \ln(x^2 y)= \frac{2xy+x^2y'}{x^2y}\]

Answer 22

that is the derivative of the first term
i assume that was the one you have trouble with?

Answer 23

Yes, I was having trouble with the derivative of ln(x^2y)

Answer 24

do you see what I did
I used product rule for the derivative of (x^2y)

Answer 25

and derivative of the ln thing is derivative of inside/inside

Answer 26

Oh, I shall work it out here in my book and see if I get that result :)

Answer 27

the result in the book?
because I didn't find y' for you
i just differentiated one term

Answer 28

I know. I meant I'll work out the derivative here in my book and see if I get the result

Answer 29

ok :p