## 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

1. hartnn Group Title

do you know the meaning of partial derivative ?

2. Babyslapmafro Group Title

yes

3. hartnn Group Title

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

4. hartnn Group Title

do you know derivative of inverse tan function ?

5. Babyslapmafro Group Title

$\frac{ 1 }{ x ^{2}+1 }$

6. hartnn Group Title

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 Group Title

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

8. hartnn Group Title

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

9. Babyslapmafro Group Title

i replaced a with 1/y

10. hartnn Group Title

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 Group Title

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

12. Babyslapmafro Group Title

numerator = (1/y)

13. hartnn Group Title

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 Group Title

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 Group Title

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

16. hartnn Group Title

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 Group Title

oh right ya, thanks

18. hartnn Group Title

welcome ^_^