#Need help
65 messages · Page 1 of 1 (latest)
<@&727457725017096242>
find du/dx and dv/dx by chain rule
Oh I see so f(x)= x^3?
no
fäf
you can get the last form easily without need of writing u in terms of v or x
got it ima give it another go
wait can you see my mistake
mistake?
in this
no
there is a mistake find it if you know the chain rule
oh ok thank you
there's only one mistake here. you just have to find it
I'm sorry I don't know it very well in the form of d/dx
just think it as normal algebra $\frac{x}{y} = \frac{x}{z}\times \frac{z}{y}$
fäf
derivation of x respect to y?
right
you can tell me then what is the correct way
i see
now you can find each derivative
and get du/dv
wait dy/dx was the right answer?
yes
oh ok tysm I will let you know if I got it
in that way dy can cancel each other
you didn't find dx/dv?
Yeah my teacher just wants us to apply the rules
but not simplify
unless he asked for it
but i think you missed dx/dv
hmmm
fäf
oh I see I did not put in v at all
Would it work if you just multiply the whole equation by 1/(2v)
oh so I can just multiply that to what I already have and I got it?
yes
so the final answer would be 3(x/(x+8))^(2)(8/(x^(2)+8)^2) x (1/2v)?
,w differentiate x/(x+8)
your middle term needs change
(x+8)^(2)?
yes
this is right
I see so if I replace it I got it right?
yeah
pls ping me when you reply
I see tysm for all the help!
or i won't see your message until i open discord
you're welcome
idk