Question

dy/dx ln(xy)=x^2-x

Answer 1

but what about the other side of the equal sign?

Answer 2

Wait, what do you have to do exactly in this problem?

Answer 3

it says consider the relation ln(xy)=xy^2-x

find dy/dx

Answer 4

Are you sure it's dy/dx ln(xy)=x²-x, not d/dx ln(xy)=x²-x?

you can just divide both sides by ln(xy) then replace it to xy²-x

Answer 5

im sure. it says:

consider the relation ln(xy)=xy^2-x
find dy/dx

Answer 6

\[\dfrac{dy}{dx} = \dfrac{x^2-x}{\ln(xy)} = \dfrac{x^2 - x}{xy^2 - x}\]Factor x out in both numerator and denominator:\[\dfrac{x(x-1)}{x(y^2-1)} = \boxed{\dfrac{x-1}{y^2-1}}\]

Answer 7

is this clear?

Answer 8

im just trying to figure out where the numbers came from is all

Answer 9

Oh and after that it says find the equation of the line tangent to the graph of the relation at the point (1,1) how do I do that?