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Mathematics 20 Online
OpenStudy (richyw):

how come if \(r=|\mathbf{x}|\) then\[\frac{\partial r}{\partial\mathbf{x}}=\frac{\mathbf{x}}{r}\]

hartnn (hartnn):

its true for every r,not just r = |x| \(r= \sqrt{x^2+y^2}\) can you diff. this w.r.t x ?

OpenStudy (richyw):

yes

OpenStudy (anonymous):

yes

hartnn (hartnn):

then tell me what u get ?

OpenStudy (richyw):

\[\frac{2x}{\sqrt{x^2+y^2}}\]

OpenStudy (richyw):

sorry openstudy is lagging again. I did that in my head and accidentaly multiplied by 2

OpenStudy (richyw):

aha I see now. thanks!

OpenStudy (richyw):

why is this true for any r?

hartnn (hartnn):

lagging for me too.... because we haven't used r =|x| . its like that is not given....

OpenStudy (richyw):

alright cool well thanks a lot!

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