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

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

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 \(z\).

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

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

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

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

You're welcome! :)

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