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Babyslapmafro Group Title

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

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    do you know the meaning of partial derivative ?

    • one year ago
  2. Babyslapmafro Group Title
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    yes

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

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

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

    • one year ago
  6. hartnn Group Title
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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?) \([\tan^{-1}{ax}]'= \dfrac{1}{1+(ax)^2}\dfrac{d}{dx}(ax)=....?\)

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

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

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

    • one year ago
  10. hartnn Group Title
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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 ?

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

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

    • one year ago
  13. hartnn Group Title
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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 :)

    • one year ago
  14. hartnn Group Title
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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 :)

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

    • one year ago
  16. hartnn Group Title
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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 : \(\large \dfrac{y}{x^2+y^2}\)

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

    • one year ago
  18. hartnn Group Title
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    welcome ^_^

    • one year ago
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