Question

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

Answer 1

do you know the meaning of partial derivative ?

Answer 2

yes

Answer 3

then have you tried to find the partial derivative ? where r u stuck ?

Answer 4

do you know derivative of inverse tan function ?

Answer 5

\[\frac{ 1 }{ x ^{2}+1 }\]

Answer 6

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?)
\([\tan^{-1}{ax}]'= \dfrac{1}{1+(ax)^2}\dfrac{d}{dx}(ax)=....?\)

Answer 7

\[\frac{ 1 }{ 1+(\frac{ 1 }{ y ^{2} }+\frac{ 2 }{ y }x+x ^{2} )}\]

Answer 8

where does the expression 1/y^2 +2y/x +x^2 come from ???

Answer 9

i replaced a with 1/y

Answer 10

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 ?

Answer 11

what about the numerator ??
d/dx (ax) =...?

Answer 12

numerator = (1/y)

Answer 13

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 :)

Answer 14

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

Answer 15

y^2/(y^2+x^2)

Answer 16

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

Answer 17

oh right ya, thanks

Answer 18

welcome ^_^