Question

Find
dy/du, du/dx, and dy/dx

y= u^2 , u=4x+7

Answer 1

2u,4,8u in order as give. :)

Answer 2

\[\frac{dy}{du}=2u\]\[\frac{du}{dx}=4\]\[\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=4\times2u=8u\]

Answer 3

can u guys show the working because i dont understand how u got that did u use the quotient rule to find dy/du

Answer 4

purple did you get and understand the solution ?

Answer 5

no

Answer 6

Just use the normal rule for differentiating polynomials to find dy/du and du/dx and the use the chain rule to find dy/dx.

Answer 7

ok

Answer 8

ok i cant do it for you
dy/dx =2u ,du/dx==4 and dy/dx= (dy/du)(du/dx)=2u(4)= 8u

Answer 9

wouldnt dy/dx be 8(4x+7)

Answer 10

i mean i can do it for u..lol..

Answer 11

using the chain rule

Answer 12

dy/dx =2u ,du/dx==4 and dy/dx= (dy/du)(du/dx)=2u(4)= 8u

Answer 13

did you understand my solution purple?

Answer 14

we didnt learn to do it like that we do f'(x) = f'(g(x))g'(x)

Answer 15

ok that is the same...lol.isnt that dy/dx is very simple?

Answer 16

nope its 8(4x+7)

Answer 17

@ least its how i learned how to do it

Answer 18

dy d(u^2) du d(4x+7)
---- = -----=2u , --- = ------ 4
du du dx dx

Answer 19

du/dx =4

Answer 20

now
dy dy du
---= --- ----= 2u(4)= 8u ans..
dx du dx

Answer 21

is it much better now purple?

Answer 22

no its the same...i'll just do that one the way i learned and see what my lectuer says

Answer 23

i didnt see that bendt did it up there..lol bec my pc keep hanging up..lol

Answer 24

but you problem reqiured dy/dx, du/dx and dy/du instead of using f(x)

Answer 25

ok

Answer 26

ok if you want to use functions
dy/du=f '(u)=2u
du/dx=f '(x)=4
dy/dx=f '(u) f '(x)= 2u(4)=8u

Answer 27

hope you get it now?..lol

Answer 28

yes

Answer 29

thank you so much