# Using implicit differentation to find the tangent to $3xy=y^2$

Use implicit differentiation to find the equation of the line tangent to the graph of $3xy = y^2$ at $(1,3)$

This is what I tried so far: \begin{align*} 3\frac{d}{dx}(x)\frac{d}{dx}(y(x)) &= \frac{d}{dx}(y(x)^2)\\ 3(1)\left(\frac{dy}{dx}\right) &= 2\frac{dy}{dx}\\ 3 &= \frac{dy}{dx} \end{align*}

Then using that to solve the equation, I get: \begin{align*} y-3 &= 3(x-1)\\ y &= 3x \end{align*}

Did I solve this correctly?

EDIT:

Ok, so I would actually have: \begin{align*} 3\left (\frac{dx}{dx}y+\frac{dy}{dx}(x)\right) &= 2y\frac{dy}{dx}\\ 3\left(y+x\left(\frac{dy}{dx}\right)\right) &= 2y\frac{dy}{dx}\\ 3\left(3+1\frac{dy}{dx}\right) &= 2(3)\frac{dy}{dx}\\ 9+3\frac{dy}{dx} &= 6\frac{dy}{dx}\\ 3 &= \frac{dy}{dx} \end{align*} Then using that to solve the equation, I get: \begin{align*} y-3 &= 3(x-1)\\ y &= 3x \end{align*}

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No. Look at the first line and remember the product and chain rules. It should be $\frac{d}{dx}(xy) = y + \frac{dy}{dx}$ and $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ –  jspecter Jun 27 '11 at 19:35
@jspecter - Question edited –  OghmaOsiris Jun 27 '11 at 19:47
You are missing a $3$ in the first line after the edit (though the omission is corrected in the second line). –  Arturo Magidin Jun 27 '11 at 19:54
And you still need to take that $3 = \frac{dy}{dx}$ and use it to answer the question asked: you want the equation of the tangent, not just the value of the derivative. –  Arturo Magidin Jun 27 '11 at 20:23
I thought that was the only thing I got right in the first draft, lol –  OghmaOsiris Jun 27 '11 at 20:24

You computed both derivatives incorrectly.

For $\frac{d}{dx}(xy)$, you need to use the product rule. You get $$\frac{d}{dx}(xy) = \left(\frac{dx}{dx}\right)y + x \left(\frac{dy}{dx}\right).$$

And for $\frac{d}{dx}y^2$, you need to use the Chain Rule. You get $$\frac{d}{dx}y^2 = 2y\left(\frac{dy}{dx}\right).$$

Once you've correctly computed the implicit derivative, you can proceed as you did before: plug in $x=1$ and $y=3$, and solve for $\frac{dy}{dx}$. That gives the slope of the tangent at $(1,3)$. Then use the point-slope formula to get the equation of the tangent.

By the way. I always tell my students to check explicitly whether the given point is actually on the given curve before starting to do computations. You might want to do that (in this case, it is).

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Question edited. –  OghmaOsiris Jun 27 '11 at 19:47
So, I fell into the right answer by mistake? Because I did that and still got $\frac{dy}{dx} = 3$ –  OghmaOsiris Jun 27 '11 at 19:59
@Oghma: Well, you solved for $y'$ incorrectly the first time (so you made lots of mistakes). If you really had $3y' = 2y'$, your conclusion should have been $y'=0$ (just subtract $2y'$ from both sides); I don't know how you managed to get $y'=3$ in the first go around. –  Arturo Magidin Jun 27 '11 at 20:03
Yeah, I dunno either.... now that I look at it, lol. –  OghmaOsiris Jun 27 '11 at 20:09