Help: Find dy/dx?

Question

Find dy/dx of:


xy^2 = ln (y) + xy^3

Answer 1

xy^2 = ln(y) + xy^3

Key thing to know when doing these type of problems is using the chain rule, and when differentiating y, express it as dy/dx.

With that said, differentiating implicitly gives us

y^2 + (1)(2y)(dy/dx) = (1/y)(dy/dx) + y^3 + x(3y^2)(dy/dx)

Move everything with a (dy/dx) to the left hand side; everything else goes to the right hand side.

(1)(2y)(dy/dx) - (1/y)(dy/dx) - x(3y^2)(dy/dx) = y^3 - y^2

2y(dy/dx) - (1/y)(dy/dx) - 3xy^2(dy/dx) = y^3 - y^2

Factor (dy/dx).

(dy/dx) (2y - (1/y) - 3xy^2) = y^3 - y^2

Let's merge into a single fraction.

(dy/dx) ( 2y^2/y - (1/y) - 3xy^3/y ) = y^3 - y^2

(dy/dx) [ 2y^2 - 1 - 3xy^3 ] / y = y^3 - y^2

Multiply both sides by y,

(dy/dx) [ 2y^2 - 1 - 3xy^3 ] = y^4 - y^3

Divide both sides to isolate dy/dx.

dy/dx = (y^4 - y^3) / (2y^2 - 1 - 3xy^3)

Answer 2

Use implicit differentiaton:

xy² = ln(y) + xy³
y² + 2xyy' = (1/y)y' + y³ + 3xy²y'
Move all terms containing y' to the LHS and all other terms to the RHS

y'(2xy - 3xy² - 1/y) = y³ - y²
y' = (y³ - y²)/(2xy - 3xy² - 1/y)
dy/dx = (y³ - y²)/(2xy - 3xy² - 1/y)
dy/dx = (y^4 - y³)/(2xy² - 3xy³ - 1)

Answer 3

ok first step will be to differentiate it all:

(power rule) (The 1 is where I differetiated x)
x(2y)[dy/dx]+y^2(1)=1/y[dy/dx]+x(3y^2)[dy/dx]+y^3(1)

now move all [dy/dx]'s to one side:

-y^3+y^2=[dy/dx][1/y+x(3y^2) - x(2y)]

get [dy/dx] alone:

[-y^3+y^2] / [y^(-1)+x(3y^2) - x(2y)]=[dy/dx]

and yeah... you can simplify the rest....

Answer 4

2xyy'+y^2 = 1/yy'+3y^2xy' +y^3
2xyy' -y'/y-3y^2xy' = y^2 + y^3
y'(2xy -1/y-3y^2x) = y^2 + y^3
y' = (y^2+y^3)/(2xy -1/y-3y^2x)