Adding extra "y" value in implicit differentiation

Question

Example

Implicit differentiate $x^2+y^2=6xy$

Textbook Answer

$2x+2y\frac{dy}{dx}=6x\frac{dy}{dx}+6y$

then textbook just says to group the $\frac{dy}{dx}$ and to solve

however, I do not understand why when you find derivative of $6xy$, you get the additional $"+6y"$. Is this a typo?

BTW

Link to problem

Before I get flagged for identical question, I wanted to make sure that the nuances of the question is actually different from this one

Answer 1

The key observation we have to make here is if $$u=6x,\quad \frac{du}{dx}=6,$$ $$v=y,\quad \frac{dv}{dx}=\frac{dy}{dx},$$ then, by virtue of the product rule $$\frac{d(6xy)}{dx}=6y+6x\frac{dy}{dx}.$$ Then, differentiating through with respect to $x$ gives $$2x+2y\frac{dy}{dx}=6y+6x\frac{dy}{dx},$$ that is $$\frac{dy}{dx}(2y-6x)=6y-2x,$$ which of course gives $$\frac{dy}{dx}=\frac{6y-2x}{2y-6x}=\frac{3y-x}{y-3x}.$$

Answer 2

Another way to look at this, which is essentially equivalent, but may help someone out there. If you look at taking a derivative, you can actually divide it up into two steps - the differential and dividing by dx. For instance, take $y = x^2$. The differential is $dy = 2x\,dx$. The derivative is found by dividing both sides by $dx$. $\frac{dy}{dx} = 2x$. Most textbooks only teach finding the derivative as a single process, rather than separating out differentiation and dividing by $dx$ to find the derivative. That's what makes your problem confusing.

If you have $6xy$, the differential is found using the product rule: $6(x\,dy + y\,dx)$. When you divide by $dx$, this becomes $6x\frac{dy}{dx} + 6y \frac{dx}{dx}$, but $\frac{dx}{dx}$ cancels itself out to 1, leaving just $6x\frac{dy}{dx} + 6y$.

Separating out the differential from the derivative simplifies the process and makes it much more understandable what is going on. For more information see here.