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Babyslapmafro

  • one year ago

Please help. How do you find the partial derivative of arctan(x/y) with respect to x?

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  1. hartnn
    • one year ago
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    do you know the meaning of partial derivative ?

  2. Babyslapmafro
    • one year ago
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    yes

  3. hartnn
    • one year ago
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    then have you tried to find the partial derivative ? where r u stuck ?

  4. hartnn
    • one year ago
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    do you know derivative of inverse tan function ?

  5. Babyslapmafro
    • one year ago
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    \[\frac{ 1 }{ x ^{2}+1 }\]

  6. hartnn
    • one year ago
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    good! now since 'y' is constant, its like taking derivative of inverse tan (ax) {where 'a' =1/y} so, apply the chain rule (know it?) \([\tan^{-1}{ax}]'= \dfrac{1}{1+(ax)^2}\dfrac{d}{dx}(ax)=....?\)

  7. Babyslapmafro
    • one year ago
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    \[\frac{ 1 }{ 1+(\frac{ 1 }{ y ^{2} }+\frac{ 2 }{ y }x+x ^{2} )}\]

  8. hartnn
    • one year ago
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    where does the expression 1/y^2 +2y/x +x^2 come from ???

  9. Babyslapmafro
    • one year ago
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    i replaced a with 1/y

  10. hartnn
    • one year ago
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    right, so, it was not (a+x)^2 , it was (ax)^2 so, the denominator would just be \(1+x^2/y^2\) got this ?

  11. hartnn
    • one year ago
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    what about the numerator ?? d/dx (ax) =...?

  12. Babyslapmafro
    • one year ago
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    numerator = (1/y)

  13. hartnn
    • one year ago
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    no....in chain rule, we need to take derivative of the inner function (here ax or x/y) which is, as you correctly said, 1/y :)

  14. hartnn
    • one year ago
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    one last step of simplification : \(\dfrac{1/y}{1+x^2/y^2}=....?\) multiply numerator and denominator by y^2 to simplify this :)

  15. Babyslapmafro
    • one year ago
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    y^2/(y^2+x^2)

  16. hartnn
    • one year ago
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    remember that the numerator was 1/y and y^2/y would just be 'y' so your final answer would look like this : \(\large \dfrac{y}{x^2+y^2}\)

  17. Babyslapmafro
    • one year ago
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    oh right ya, thanks

  18. hartnn
    • one year ago
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    welcome ^_^

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