Calculus partial derivatives problem [y^(-3/2)arctan(x/y)] * *

[jawadur_bd]

Calculus partial derivatives problem [y^(-3/2)arctan(x/y)] *urgent*

Homework Statement

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].

Homework Equations

The Attempt at a Solution

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

 

[mighty2000]

Hello there!

Let's start by looking at the product rule you used for fx(x,y):

fx(x,y) = (y^(-1/2)) * (1/(y^2+x^2)) + (y^(-3/2)) * (1/y)

Notice that in the second term, you took the derivative of arctan(x/y) with respect to x, which is correct. However, in the first term, you took the derivative of y^(-3/2) with respect to x, which is incorrect. Remember, when we use the product rule, we only take the derivative of the first term with respect to the variable we are differentiating with respect to. So in this case, we should be taking the derivative of y^(-1/2) with respect to x, which gives us (-1/2)y^(-3/2).

Therefore, the correct answer for fx(x,y) is:

fx(x,y) = -(1/2)y^(-3/2) * (1/(y^2+x^2)) + (y^(-3/2)) * (1/y)

= -y^(-3/2) * (1/(2(y^2+x^2))) + (y^(-3/2)) * (1/y)

= y^(-3/2) * (1/y - 1/(2(y^2+x^2)))

= y^(-3/2) * (2(y^2+x^2) - y)/(2y(y^2+x^2))

= y^(-3/2) * (2y^2 + 2x^2 - y)/(2y(y^2+x^2))

= (2y^(-1/2) + 2x^2y^(-3/2) - y^(-5/2))/(2(y^2+x^2))

For fy(x,y), you did everything correctly except for one small mistake. When taking the derivative of y^(-3/2) with respect to y, you forgot to use the chain rule. Remember, when we have a function inside another function, we need to use the chain rule. So the correct answer for fy(x,y) is:

fy(x,y) = -(3/2)y^(-5/2) * (1/(y^2+x^2)) - (y^(-3/2