Question

Calculus problem? http://i.imgur.com/ymoDl.jpg

Answer 1

Do I have to use the quotient rule here?

Answer 2

you can always avoid using the quotient rule if you use the chain rule+product rule instead, but that is one way to do it

Answer 3

take the derivative of z(x) using the defintion they gave you and see what you get

Answer 4

@TuringTest I got z + dz/dx

Answer 5

No wait z + xdz/dx

Answer 6

Done using the product rule y=zx

Answer 7

@ironictoaster do you get it?

Answer 8

I assume I move the z to the RHS?

Answer 9

I'm having a bit of trouble with it myself

Answer 10

Actually I have no idea.

Answer 11

\[z(x)=\frac {y(x)}x\implies z'={y'x-y\over x^2}\]\[xz'={y'x-y\over x}=y'-\frac yx=y'-z\]but I don't know what to sub for y' to get it in terms of z

Answer 12

@TuringTest Maybe the question before it will help you make any sense of it.

Answer 13

Although part a is very straight forward.

Answer 14

yeah I needed the rest of the info

Answer 15

Aw crap, sorry!

Answer 16

it's ok, lol
we have to sub for y' like I said, and the previous line says that \[y'=(\frac yx)^2+\frac yx-1\]if\[z=\frac yx\]then\[y'=z^2+z-1\]sub that into what I got above for xz' and you have the answer

Answer 17

What about the + z from the product rule?

Answer 18

\[z(x)=\frac {y(x)}x\implies z'={y'x-y\over x^2}\]\[xz'={y'x-y\over x}=y'-\frac yx=y'-z\]and from the previous line we have that\[y'=(\frac yx)^2+\frac yx-1=z^2+z-1\]so we sub that in to get\[xz'=z^2+\cancel z-1-\cancel z\]

Answer 19

Hmm, I didn't use the quotient rule at all..

Answer 20

then use the product rule
write it as\[z(x)=y(x)x^{-1}\]

Answer 21

There's so many x's and y's and z's I am confusing myself..!

Answer 22

we know what y' is

Answer 23

we are told that z=y/x , and that both z and y are functions of x
find the derivative of z using the product rule or however you prefer, then plug in the expression for y' you have from the previous part