#Partial Derv, Part C?
33 messages · Page 1 of 1 (latest)
you would take the derivative relative to x, and then the derivative relative to y of that:
∂/∂y(∂f/∂x)
and if the function is continuous, as well as the partial derivative relative to x, and the partial derivative relative to y, then that should be the same as ∂/∂x(∂f/∂y)
@lost oyster
the detail I added about continuity links into mean value theorem, when you get to real analysis then you'll learn about how it works and why it's related with calc
ohhh
np
I put it into a calc and it gave me -6x? But when I did it on paper I got 8x^3 -6x
It's the 3rd line
From the top
one sec:
first derivative relative to x:
(-2x)(y-2x²)-(4x)(y-x²)
derivative relative to y:
-2x-4x
I used product rule first
I go that for the first derv, then simplified down to 8x^3-6xy
-2xy+4x³-4xy+4x²
8x³-6xy
and then relative to y
8x³ is constant, so is 0
-6xy→-6x
wait
what
now you have ∂³f/∂x∂y²
OHHH
yes lmao
I wasn't using the given f(x,y) lmao