Partial derivative of f(x,y): x*y^-1 + y*z^-1 +z*x^-1 wrt x? Wouldn't it just be 1/y + z/x^2 ?

Question

anonymous · Answer

close, your last term is missing a negative.

tiffany_rhodes · Answer

Why would the last term be negative?

anonymous · Answer

what's the derivative of lets say 2\x?

anonymous · Answer

with respect to x of course.

tiffany_rhodes · Answer

2/x^2 ?

anonymous · Answer

no, it's - 2/x^2

anonymous · Answer

because the exponent of x is really -1 before the derivative right?

tiffany_rhodes · Answer

Oh okay. I get it.

tiffany_rhodes · Answer

so basically 2*x^-1 --> 2*-1x^(-2)

tiffany_rhodes · Answer

so -2/x^2

anonymous · Answer

yes exactly!

tiffany_rhodes · Answer

My calculus is a little rusty lol.

anonymous · Answer

tell me about it, I haven't done math in over 5 years lol.

tiffany_rhodes · Answer

Oh wow, you don't seem to rusty though :) So partial derivative wrt x would just be 1/y + -z/x^2 ?

tiffany_rhodes · Answer

too*

anonymous · Answer

yes! But trust me it is, I went as high as linear algebra and I will not be able to help you do anything with matrices...or anything too complex.

tiffany_rhodes · Answer

Yeah, I took linear algebra and the math was easy but conceptually it was difficult. Thanks for the help though! :)

anonymous · Answer

no problem.