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Mathematics 81 Online
OpenStudy (anonymous):

let f(x,y)=x^2y+y^2z. use chain rule to calculate df/ds, where x = s+t, y = st, z = 2s-t. the answer if got is df/ds=2xy+x^2t+2yzt+2y^2s i am working out a test review and need to know if thats right! let f(x,y)=x^2y+y^2z. use chain rule to calculate df/ds, where x = s+t, y = st, z = 2s-t. the answer if got is df/ds=2xy+x^2t+2yzt+2y^2s i am working out a test review and need to know if thats right! @Mathematics

OpenStudy (anonymous):

\[{\partial f \over \partial s}={\partial f \over \partial x}{\partial x \over \partial s}+{\partial f \over \partial y}{\partial y \over \partial s}+{\partial f \over \partial z}{\partial z \over \partial s}=2xy(1)+(x^2+2yz)(t)+y^2(2).\] Simplifying yields to \(\frac{\partial f}{\partial s}=2xy+x^2t+2yzt+2y^2.\)

OpenStudy (anonymous):

The only difference from your answer is that you have an s in your last term. This is wrong unless you made a mistake in your question regarding \(z\).

OpenStudy (anonymous):

ok yeah i messed up dz/ds = 2s instead of just 2, which is why i had the extra s at the end.

OpenStudy (anonymous):

but thanks again AnwaraA! by the way, how do you make the partial derivative symbol, are you using another program to type equations?

OpenStudy (anonymous):

No. You can use the "Equation" button below and write "partial". Call me Anwar!

OpenStudy (anonymous):

oh ok thats pretty cool. thanks again Anwar!:)

OpenStudy (anonymous):

You're welcome! :)

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