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

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do you know the meaning of partial derivative ?
then have you tried to find the partial derivative ? where r u stuck ?

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Other answers:

do you know derivative of inverse tan function ?
\[\frac{ 1 }{ x ^{2}+1 }\]
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?) \([\tan^{-1}{ax}]'= \dfrac{1}{1+(ax)^2}\dfrac{d}{dx}(ax)=....?\)
\[\frac{ 1 }{ 1+(\frac{ 1 }{ y ^{2} }+\frac{ 2 }{ y }x+x ^{2} )}\]
where does the expression 1/y^2 +2y/x +x^2 come from ???
i replaced a with 1/y
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 ?
what about the numerator ?? d/dx (ax) =...?
numerator = (1/y) 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 last step of simplification : \(\dfrac{1/y}{1+x^2/y^2}=....?\) multiply numerator and denominator by y^2 to simplify this :)
remember that the numerator was 1/y and y^2/y would just be 'y' so your final answer would look like this : \(\large \dfrac{y}{x^2+y^2}\)
oh right ya, thanks
welcome ^_^

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