Guest wrote: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].
It might be simpler to set x= 2, y= 1 here:
[tex]1+ 4\frac{dy}{dx}+ 3- 3\frac{dy}{dx}= 0[/tex]
[tex]4+ \frac{dy}{dx}= 0[/tex] so [tex]\frac{dy}{dx}= -4[/tex].
You will notice that is not the same as my previous answer. I had, to my embarrassment, swapped a "-" to a "+".
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].
No, it should be [tex]\frac{dy}{dx}= \frac{1^2+ 3}{3- 2(1)(2)}= \frac{4}{-1}= -4[/tex].