Calculus partial derivatives problem [f(x,y) = y^(-3/2)arctan(x/y)...find fx(x,y) and fy(x,y)] ?

Question

mathematics is not my strong suit..i tried the problem from a couple of different angles..i am not getting the correct answer

here is the problem:

f(x,y) = y^(-3/2)arctan(x/y)...find fx(x,y) and fy(x,y) [as in derivatives with respect to x and with respect to y].

here is what i have tried doing so far..i used the product rule obviously. we know that derivative of (1/a)arctan(x/a) gives (1/(a^2+x^2)

i took y^(-3/2) as 'u' and arctan(x/y) as 'v' for the implementation of the product rule. so am i getting y/(y^2+x^2) as the derivative (since (1/y) is missing from the arctan term)?

for fx(x,y) i get the answer y^(-1/2)/(y^2+x^2)..this answer matches with the book's answer (ch-13.3 prob no.25 - calculus 9th ed by anton, bivens, davis)

however i am not sure if i got it correct only by chance since i used the same method for fy(x,y) only to get an incorrect answer..my answer for fy(x,y) came -(3/2)y^(-5/2)arctan(x/y)-(xy^(-5/2))/(y^2+x^2)..

the correct answer is -(3/2)y^(-3/2)arctan(x/y)-(xy^(-3/2))/(y^2+x^2).....[note: (xy^(-3/2)...not (xy^(-5/2)]

fy(x,y) = -(3/2)(y^(-5/2))arctan(x/y)+(y^(-3/2))(y/(y^2+x^2))(-x/(y^2))...which gives: -(3/2)y^(-5/2)arctan(x/y)-(xy^(-5/2))/(y^2+x^2)..so what went wrong there..?

been stuck for hours its quite frustrating....so can anyone please show me the workings with the steps so that i know where i am getting it wrong?

thanks in advance! :)

Answer 1

f(x)=y^-(3/2)arctan(x/y).

Consider z=arctan(x/y), then tan(z)=x/y, and sec²(z)∂z/∂x=1/y.

sec²(z)≡1+tan²(z)=1+(x/y)², ∂z/∂x=(1/y)/(1+(x/y)²)=1/(y+x²/y)=y/(x²+y²).

sec²(z)∂z/∂y=-x/y², ∂z/∂y=-(x/y²)/(1+(x/y)²)=-x/(x²+y²).

Consider w=y^-(3/2), ∂w/∂y=-(3/2)y^-(5/2). ∂w/∂x=0.

f(x)=wz,

fᵪ=∂f/∂x=w∂z/∂x+z∂w/∂x=(y^-(3/2))/(y+x²/y) or (y^(-1/2))/(x²+y²).

fᵧ=∂f/∂y=w∂z/∂y+z∂w/∂y=-x(y^-(3/2))/(x²+y²)-((3/2)y^-(5/2))arctan(x/y).