What is the derivative of
2x^2-xy+y^2-x-5y+8 with respect to x....this is th only problem that is holding me back.
2006-10-09 09:26:51 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
d/dx(2x^2 -xy -x+y^2-5y + 8)
d/dx(2x^2)+ d/dx(-xy) + d/dx(-x) +d/dx(y^2) +d/dx(-5y) +d/dx(8)
4x -y -5d/dx(y) + d/dx (y^2) -1
4x - y + d/dx (y^2) -1
The derivative of the constanty^2 is zero
hence,
4x-y-1
This is the answer
2006-10-09 10:04:36 · answer #1 · answered by quark_sa 2 · 0⤊ 0⤋
Simply take the derivative wrt x as you normally would and remember that dx/dx is 1. For the first term,
d/dx 2x² = 4x dx/dx = 4x For the next term, remember the product rule:
d/dx (-xy) = -(x dy/dx + y dx/dx) = -x dy/dx - y
Now *you* do the rest of it âº
Doug
2006-10-09 16:39:38 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
y = 2x^2-xy+y^2-x-5y+8
dy/dx = 4x -x*dy/dx - y + 2y*dy/dx - 1 - 5*dy/dx
2006-10-09 16:48:28 · answer #3 · answered by عبد الله (ドラゴン) 5 · 0⤊ 0⤋
you have to treat y as a constant
4x -y -1
2006-10-09 16:29:05 · answer #4 · answered by kevvsworld 3 · 2⤊ 0⤋