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quick differential question (3 terms)

y = xz

therefore dy/dx = x(dz/dx) + z

can anybody explain why this is? lol

EDIT: nvm.. i just remembered from last year that dy/dx = dy/dz x dz/dx
(edited 13 years ago)
Original post by ucasisannoying
y = xz

therefore dy/dx = x(dz/dx) + z

can anybody explain why this is? lol

EDIT: nvm.. i just remembered from last year that dy/dx = dy/dz x dz/dx

This has nothing to do with the chain rule. This is done using the product rule.
ddx[xz(x)]=xddx[z(x)]+z(x)ddx[x]=xdzdx+z\frac{d}{dx}[xz(x)] = x\frac{d}{dx}[z(x)] + z(x)\frac{d}{dx}[x] = x\frac{dz}{dx} + z.
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Avatar for ucasisannoying
ucasisannoying
OP
Original post by Farhan.Hanif93
This has nothing to do with the chain rule. This is done using the product rule.
ddx[xz(x)]=xddx[z(x)]+z(x)ddx[x]=xdzdx+z\frac{d}{dx}[xz(x)] = x\frac{d}{dx}[z(x)] + z(x)\frac{d}{dx}[x] = x\frac{dz}{dx} + z.
oh damn... works for both rules though, what exactly is stopping us from using the chain rule here? i thought the principle that dy/dx = dy/du x du/dx where u is a variable works regardless
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Avatar for Kevlar
Kevlar
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Original post by ucasisannoying

Original post by ucasisannoying
oh damn... works for both rules though, what exactly is stopping us from using the chain rule here? i thought the principle that dy/dx = dy/du x du/dx where u is a variable works regardless


Chain rule won't work here, with 2 variables think you'd take partials differentials for it to work, instead this is simply reverse product rule,
Original post by Kevlar
...this is simply reverse product rule

What is this, I don't even... :lolwut:

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