by **Guest** » Fri Oct 16, 2020 11:30 am

I assume that the "y2" is supposed to be [tex]y^2[/tex] (a better notation is "y^2").

The function is [tex]xy^2+ 3x- 3y= 6[/tex]. To find the derivative of y without having to solve for y, use the "chain rule". The derivative of [tex]xy^2[/tex] with respect to x is [tex]1(y^2)+ x\frac{d(y^2)}{dy}\frac{dy}{dx}= y^2+ 2xy\frac{dy}{dx}=0[/tex]. Differentiating [tex]xy^2+ 3x- 3y= 6[/tex] with respect to x gives [tex]y^2+ 2xy\frac{dy}{dx}+ 3- 3\frac{dy}{dx}= 0[/tex].

Now solve for dy/dx: [tex](2xy- 3)\frac{dy}{dx}= -3- y^2[/tex]. [tex]\frac{dy}{dx}= \frac{y^2+ 3}{3- 2xy}[/tex].

Setting x= 2, y= 1, [tex]\frac{dy}{dx}= \frac{1^2+ 3}{3+ 2(1)(2)}= \frac{4}{7}[/tex].