Calculus: Math 250: Implicit differentiation?

Question

Alright, the instructions are to find dy/dx by implicit differentiation for the following problems:

(x^2)y+x(y^2)=3x
When I work this problem, I get y'=3/(2x+2y) The back of the book says y'=(3-2xy-y^2/(x^2+2xy)

(x^3)+[(x^2)y]+4y^2=6
Once again, I get y'=(-3x^2)/(2x+8y) when the book says I should be getting y'=[-x(3x+2y)]/(x^2+8y)

and one more example: (x^2)(y^2)+xsiny=4
Book says y'=(-2x(y^2)-siny)/(2(x^2)y+xcosy)

Help would be greatly appreciated. Since I'm consistantly wrong, I think maybe I'm just starting off with the wrong concepts. If you do answer my question, could you please try to explain each step? I don't just want the answers; I'm trying really hard to learn this.

Thanks!

Answer 1

You are not muliplying it out.
you have to use the multiplication rule uv'+vu'
For example: the derivative of x^2y= 2xy+x^2y'

Answer 2

when you differentiate the first one you should get:

2xy + (x^2)y' + y^2 + 2xyy' = 3 and when you solve for y' you get the right answer.

well you're differentiating with respect to x, so every variable that is not x, like y, you tag on a y' to it. Also you need to do the product rule in this problem so for (x^2)y, you take the derivative of the first (x^2) times the second normally (y), so you get (2x * y) and then add (+) the first (x^2) times the derivative of the second (y) so (x^2) *(1 * y') and its
2xy +(x^2)y'

Answer 3

(x^2)y+x(y^2)=3x
Using the product rule,
(x^2)dy + 2xydx + 2xydy + (y^2)dx = 3dx
(x^2 + 2x)dy = (3 - 2xy - y^2)dx
dy/dx = (3 - 2xy - y^2)/(x^2 + 2x)

Again,
(x^3) + [(x^2)y] + 4y^2 = 6
3x^2dx + [2xydx + x^2dy] + 8ydy = 0
(x^2 + 8y)dy = -(3x^2 + 2xy)dx
dy/dx = - x(3x + 2y)/(x^2 + 8y)

[(x^2)(y^2)] + {xsiny} = 4
[(x^2)2ydy + (y^2)2xdx + {xcosydy + (siny)dx} = 0
(2yx^2 + xcosy)dy = - (2xy^2 + siny)dx
dy/dx = - (2xy^2 + siny)/(2(x^2)y + xcosy)

Answer 4

answering 3 questions will be long so let me work out one of thern

we are given
x^2y+xy^2 = 3x
differentiate both sides wrt x

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

now d/dx(x^2y+xy^2)
= d/dx(x^2y) + d/dx(xy^2)

using product rule(d/dx(vu) = udv/dx + vdu/dx)
d/dx(x^2y) = x^2 dy/dx + y d/dx(x^2) = x^2dy/dx + 2xy
d/dx(xy^2) = x2ydy/dx + y^2
now we get
x^2dy/dx + 2xy+ 2xydy/dx +y^2 = 3

or dy/dx(x^2+2xy) = (3-y^2-2xy)
dy/dx = (3-y^2-2xy)/(x^2+2xy)
so back of the book is correct

similarly you can work out the other 2
.