Use implicit differentiation to find the derivative of y with reapect to x
x^2y-y^2+ln(xy)=1
x^2 dy+2xydx - 2ydy +(1/xy)(xdy+ydx) = 0
[x^2-2y+1/y]dy + [2xy+1/x]dx = 0
dy/dx = -[2xy+1/x] / [x^2-2y+1/y]
Or, rearranging things a bit,
-y(2x^2y+1) / x(x^2y-2y^2+1)
= -y(2x^2y+1) / x(x^2y-y^2-1 + 2-y^2)
= y(2x^2y+1) / x(2-y^2+ln(xy))
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To find the derivative of y with respect to x using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule.
For the left-hand side of the equation:
- The derivative of x^2y with respect to x is 2xy + x^2(dy/dx).
- The derivative of y^2 with respect to x is 2yy'.
- The derivative of ln(xy) with respect to x requires the use of the chain rule, which states that the derivative of ln(u) with respect to x is (1/u) * (du/dx). In this case, u = xy, so the derivative is (1/(xy)) * (d(xy)/dx) = (1/(xy)) * (y + x(dy/dx)).
For the right-hand side of the equation, the derivative of 1 with respect to x is 0.
Step 2: Simplify the resulting equation by collecting all terms involving dy/dx on one side, and all other terms on the other side.
The resulting equation after differentiation is:
2xy + x^2(dy/dx) - 2yy' + (1/(xy)) * (y + x(dy/dx)) = 0.
Step 3: Solve the resulting equation for dy/dx to obtain the derivative of y with respect to x.
First, simplify the equation by combining like terms:
x^2(dy/dx) + (1/(xy))(x(dy/dx)) - 2yy' + (2xy + y) = 0.
Now, isolate dy/dx:
x^2(dy/dx) + (1/(xy))(x(dy/dx)) - 2yy' = -2xy - y.
Factor out the common factor of dy/dx:
dy/dx(x^2 + 1/(xy)) - 2yy' = -2xy - y.
Finally, solve for dy/dx:
dy/dx = (-2xy - y) / (x^2 + 1/(xy)).
Therefore, the derivative of y with respect to x is given by (-2xy - y) / (x^2 + 1/(xy)).