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    For\ f(x,y)=ln\sqrt{x^2+y^2},\ show\ that\ x\frac{\delta f}{\delta x}+y\frac{\delta f}{\delta y}=1

    Advice would be appreciated
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    I've not been taught partial differentiation properly but you treat the other variables as constants right?
    Then use the chain rule to get
     \frac{\delta}{\delta x} (ln \sqrt{x^2 + y^2}) =  \frac{\delta}{\delta x}(\sqrt{x^2 + y^2})  . \frac{1}{\sqrt{x^2 + y^2}}
    Works similarly for df/dy.
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    (Original post by vector)
    For\ f(x,y)=ln\sqrt{x^2+y^2},\ show\ that\ x\frac{\delta f}{\delta x}+y\frac{\delta f}{\delta y}=1

    Advice would be appreciated
    You may find it easier to write

    ln\sqrt{x^2+y^2}\text{  as  }\frac{1}{2}ln(x^2+y^2)

    before finding your derivaties.
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    (Original post by benwellsday)
    I've not been taught partial differentiation properly but you treat the other variables as constants right?
    Then use the chain rule to get
     \frac{\delta}{\delta x} (ln \sqrt{x^2 + y^2}) =  \frac{\delta}{\delta x}(\sqrt{x^2 + y^2})  . \frac{1}{\sqrt{x^2 + y^2}}
    Works similarly for df/dy.
    erm d/dx [f(x)] = f'(x)/f(x) :p:

    So  \frac{\partial}{\partial x} (ln \sqrt{x^2 + y^2}) = [x(x^2 + y^2)^{-1/2}]\frac{1}{\sqrt{x^2 + y^2}} = \frac{x}{x^2 + y^2}
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    (Original post by EierVonSatan)
    erm d/dx [f(x)] = f'(x)/f(x) :p:

    So  \frac{\partial}{\partial x} (ln \sqrt{x^2 + y^2}) = [x(x^2 + y^2)^{-1/2}]\frac{1}{\sqrt{x^2 + y^2}} = \frac{x}{x^2 + y^2}
    Which is what I got, but I tried not to give the answer away!
    I probably didn't write it out in a great way but I meant do   \frac{1}{\sqrt{x^2 + y^2}} \times \frac{\partial}{\partial x} (\sqrt{x^2 + y^2})
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    (Original post by benwellsday)
    Which is what I got, but I tried not to give the answer away!
    I probably didn't write it out in a great way but I meant do   \frac{1}{\sqrt{x^2 + y^2}} \times \frac{\partial}{\partial x} (\sqrt{x^2 + y^2})
    lol ahh I see I misread, sorry :o:
 
 
 
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