Find dy/du, du/dx, and dy/dx y= u^2 , u=4x+7

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Find dy/du, du/dx, and dy/dx y= u^2 , u=4x+7

Mathematics
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2u,4,8u in order as give. :)
\[\frac{dy}{du}=2u\]\[\frac{du}{dx}=4\]\[\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=4\times2u=8u\]
can u guys show the working because i dont understand how u got that did u use the quotient rule to find dy/du

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Other answers:

purple did you get and understand the solution ?
no
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.
ok
ok i cant do it for you dy/dx =2u ,du/dx==4 and dy/dx= (dy/du)(du/dx)=2u(4)= 8u
wouldnt dy/dx be 8(4x+7)
i mean i can do it for u..lol..
using the chain rule
dy/dx =2u ,du/dx==4 and dy/dx= (dy/du)(du/dx)=2u(4)= 8u
did you understand my solution purple?
we didnt learn to do it like that we do f'(x) = f'(g(x))g'(x)
ok that is the same...lol.isnt that dy/dx is very simple?
nope its 8(4x+7)
@ least its how i learned how to do it
dy d(u^2) du d(4x+7) ---- = -----=2u , --- = ------ 4 du du dx dx
du/dx =4
now dy dy du ---= --- ----= 2u(4)= 8u ans.. dx du dx
is it much better now purple?
no its the same...i'll just do that one the way i learned and see what my lectuer says
i didnt see that bendt did it up there..lol bec my pc keep hanging up..lol
but you problem reqiured dy/dx, du/dx and dy/du instead of using f(x)
ok
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
hope you get it now?..lol
yes
thank you so much

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