## Calculus, 10th Edition (Anton)

$y' = \dfrac{10xy-3x^2y^2-1}{2x^3y-5x^2}$
In order to derivate this function you have to apply implicit differentation method. First, take the function to it's f(x)=0 form $x^3y^2-5x^2y+x-1=0$ Then derivate the whole equation. Rember the put y' every time you derivate y $3x^2y^2+2x^3yy'-10xy-5x^2y'+1=0$ *Note: Here you have to apply the product rule twice Solve for y' and you have the answer $y'(2x^3y-5x^2) =10xy-3x^2y^2-1$ $y' = \dfrac{10xy-3x^2y^2-1}{2x^3y-5x^2}$