Calculus Question

[mrkun]

What are the partial derivatives f(x,y) = arccos (xy) with respect to x and with respect to y?

My answers were:

az/ax = -y/[1-(y^2)(x^2)] and az/ay = -x/[1-(y^2)(x^2)]

However, I'm supposed do it at the point (1,1) at which they both equal -1/0. So, I'm guessing I'm taking the derivative incorrectly.

 

[Random Variable]

you forgot the square root sign

but the derivative still doesn't exist at (1,1)

 

[mrkun]

Yeah, you're right, it should be

az/ax = -y/[1-(y^2)(x^2)]^(1/2) and az/ay = -x/[1-(y^2)(x^2)]^(1/2)

I had that originally, just forgot to type it. But, I still thought I'd made an error. Anyone else?