# Math Help - Slope of the tangent

1. ## Slope of the tangent

The slope of the tangent to the curve $y^3x + y^2x^2 = 6$ at (2,1) is:

$-3/2, -1, -5/14, -3/14, 0$

What am I suppose to do? Implicit differentiation? I tried that and I ended up getting $3y' + 8y' = 0$

What do I do??

2. Originally Posted by Reefer
The slope of the tangent to the curve $y^3x + y^2x^2 = 6$ at (2,1) is:

$-3/2, -1, -5/14, -3/14, 0$

What am I suppose to do? Implicit differentiation? I tried that and I ended up getting $3y' + 8y' = 0$

What do I do??
$y^3x+y^2x^2=6$

Note in the following explanation $\frac{dy}{dx}=y'$. I prefer using $\frac{dy}{dx}$ instead of $y'$. Old habits die hard...

Implicit differentiation yields:

$3y^2\frac{dy}{dx}(x)+y^3(1)+2y\frac{dy}{dx}(x^2)+y ^2(2x)=0$

(remember you have to use product rule...x and y are both variables...so when differentiating the product we have to use product rule)

We want all the terms with $\frac{dy}{dx}$ on one side so we can factor out that $\frac{dy}{dx}$ and solve for it. So the previous equation turns into:

$3y^2(x)\frac{dy}{dx}+2y(x^2)\frac{dy}{dx}=-y^3-2xy^2$

Factoring out $\frac{dy}{dx}$ on the left hand side yields:

$\frac{dy}{dx}(3y^2x+2yx^2)=-y^3-2xy^2$

Now, to get $\frac{dy}{dx}$ by itself, we must divide through by $3y^2x+2yx^2$ which results in:

$\frac{dy}{dx}=y'=\frac{-y^3-2xy^2}{3y^2x+2yx^2}$

You want to find the slope of the tangent line at (x,y)=(2,1). We have solved for $\frac{dy}{dx}$ which gives the slope of the tangent line at any point on the original function. So now, you just need to substitute x=2 and y=1 into the equation for $\frac{dy}{dx}$.

Does that make sense?

3. Oh I get it.. Didn't know we had to use the product rule.

4. Well, it's good that you ran across that in your homework as opposed to an in-class exam! I figured that was the issue..I differentiated it disregarding the x and came up with the same thing you did.