Find (dy/dx)?

Question

Find (dy/dx) for the equation (x^3)-2(x^2)(y)+3x(y^2)=38

Answer 1

d/dx (x³ - 2 x² y + (3x)(y²) ) = 0
3x² - 4xy - 2x² dy/dx + 3y² + 2y dy/dx (3x) = 0
3x² - 4 xy - 2x² dy/dx + 3y² + 6xy dy/dx = 0
3x² - 4xy + 3y² + ( 6xy - 2x² ) dy/dx = 0
3x² - 4xy + 3y² + (2x)(3y - x) dy/dx = 0
dy/dx = ( 4xy - 3x² - 3y² ) / [ (2x) (3y - x) ]

Answer 2

In differentiation, the derivative of y = dy, and that of x = dx...
so technically, when u say that u want the derivative of:
y = x thus it is dy = dx, and u divide the dy by the dx, so that the equation will become dy/dx = 1
notice that 1 is the derivative of x when solving equations normally!!!!
so u can either use that procedure, or, u can use a simpler one which is that u differentiate the x normally like always, and when it comes to differentiating y, the derivative is dy/dx:

(x^3) - 2(x^2)(y) +3x(y^2) = 38
so u differentiate the whole equation:
by using the product rule:
d 2(x^2)(y) = 4x(y) + 2(x^2)dy/dx

and d(3x (y^2)) = 3(y^2) + 3x(2y)dy/dx

so d(the equation) = 3(x^2) - { 4xy + 2(x^2)dy/dx } +3(y^2) +(6xy)dy/dx = 0

3x^2 -4xy -2x^2(dy/dx) +3y^2 +6xy(dy/dx) = 0
4xy -3x^2 -3y^2 = -2x^2(dy/dx) +6xy(dy/dx)
4xy -3x^2 -3y^2 = (dy/dx) (6xy - 2x^2)
so dy/dx = (4xy -3x^2 -3y^2) / (6xy -2x^2)

Answer 3

Find (dy/dx) for the equation Find (dy/dx) for the equation (x^3)-2(x^2)(y)+3x(y^2)=38

Solution:
(x^3)-2(x^2)(y)+3x(y^2)=38
By Implicit differentiation:
dy/dx = 3x^2 - 2[2xy + x^2dy/dx] + 3[1(y^2) + 2yxdy/dx} = 0
3x^2 - 4xy - 2x^2dy/dx + 3y^2 + 6xydy/dx = 0
dy/dx(-2x^2 + 6xy) = -3x^2 + 4xy - 3y^2
dy/dx = (-3x^2 + 4xy - 3y^2)/(-2x^2 + 6xy) ANS

teddy boy

Answer 4

assuming y=f(x),

d(2 x^2 y)/dx
= d(2x^2)/dx y + 2x^2 dy/dx (product rule)
= 4x^2 y + 2x^2 dy/dx

and

d(3x y^2 )/dx
= d(3x)/dx y^2 + 3x d(y^2)/dx (product rule)
= 3 y^2 + 3x 2y dy/dx (chain rule)
= 3y^2 + 6xy dy/dx
so we have

d( x^3 - 2 x^2 y + 3x y^2 )/dx = d( 38 )/dx
3x^2 - 4xy - 2x^2 dy/dx + 3y^2 + 6xy dy/dx = 0
(6xy - 2x^2) dy/dx = 4xy - 3x^2 - 3y^2
dy/dx = (4xy - 3x^2 - 3y^2)/(6xy - 2x^2)

Answer 5

(d/dx)[(x^3)-2(x^2)(y)+3x(y^2)] = (d/dx)(38)

3x^2 - 4xy - 2x^2(dy/dx) + 3y^2 + 3x(2y(dy/dx)) = 0

3x^2 - 4xy + 3y^2 -(2x^2 - 6xy)(dy/dx) = 0

3x^2 - 4xy + 3y^2 = (2x^2 - 6xy)(dy/dx)

(3x^2 - 4xy + 3y^2)/(2x^2 - 6xy) = (dy/dx)