# Finding equation of the tangent to the curve

### Finding equation of the tangent to the curve

I need to find the equation of the tangent to the curve xy2 + 3x -3y=6. at the point (2,1)

I know i need to differentiate implicitly but its getting confusing differentiating the xy2?
After differentiation then id set the function dy/dx to equal 0???
Sylus

### Re: Finding equation of the tangent to the curve

I assume that the "y2" is supposed to be $$y^2$$ (a better notation is "y^2").
The function is $$xy^2+ 3x- 3y= 6$$. To find the derivative of y without having to solve for y, use the "chain rule". The derivative of $$xy^2$$ with respect to x is $$1(y^2)+ x\frac{d(y^2)}{dy}\frac{dy}{dx}= y^2+ 2xy\frac{dy}{dx}=0$$. Differentiating $$xy^2+ 3x- 3y= 6$$ with respect to x gives $$y^2+ 2xy\frac{dy}{dx}+ 3- 3\frac{dy}{dx}= 0$$.

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

Setting x= 2, y= 1, $$\frac{dy}{dx}= \frac{1^2+ 3}{3+ 2(1)(2)}= \frac{4}{7}$$.
Guest