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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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    I thought that was the only thing I got right in the first draft, lol2011-06-27

1 Answers 1

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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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    Yeah, I dunno either.... now that I look at it, lol.2011-06-27