Equations: 3x-5y = 16 and xy = 7 Solutions: (7, 1) and (-5/3, -21/5)
The differential of the product xy with respect to x is y + x dy/dx. The differential of logy with respect to x is (1/y) dy/dx. The role of c in this question is not made clear.
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
mid point of xy
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
Equations: 3x-5y = 16 and xy = 7 Solutions: (7, 1) and (-5/3, -21/5)
The differential of the product xy with respect to x is y + x dy/dx. The differential of logy with respect to x is (1/y) dy/dx. The role of c in this question is not made clear.
A system of linear equations determines a line on the xy-plane. The solution to a linear set must satisfy all equations. The solution set is the intersection of x and y, and is either a line, a single point, or the empty set.
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
xy = x ÷x y = 1
xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y
That is the commutative property of equality.
mid point of xy
This gives us the equations: n = xy = yx x != y (Note that for this example, the != stands for not equal to similar to some programming languages.) xy always equals yx due to the communtative property of multiplication. So there are actually an infinite number of answers. Some are given below: 6 = (2)(3) = (3)(2), 2 != 3 12 = (3)(4) = (4)(3), 3 != 4
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
18
Algebraic expressions have derivatives. Equations have solutions (sometimes). I suppose you could dfferentiate each term of this equation with respect to x: cos(xy) gives -sin(xy)(y+xdy/dx), -y3 gives -3y2dy/dx, and 4x2y gives 8xy +4x2dy/dx