Babyslapmafro
  • Babyslapmafro
Please help. How do you find the partial derivative of arctan(x/y) with respect to x?
Mathematics
katieb
  • katieb
See more answers at brainly.com
katieb
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.