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\[y^4+3y-4x^3=5x+1\]
The answer was \[y^4y'+3y'-12x^2=5\]

The one thing that confuses me is the y's. Are you supposed to add another y variable?

\[ \frac{d}{dx} y^4= 4 y^3 \frac{d}{dx} y \\
4y^3 \frac{dy}{dx} \]

another example
\[ -4 \frac{d}{dx} x^3= -12 x^2 \frac{dx}{dx} \\ =-12x^2 \]

Is there a reason why it goes away for x and not y?

I think of it like this:
\[ \frac{d}{dx} x = \frac{dx}{dx} = 1\\ \frac{d}{dx} y= \frac{dy}{dx} \]

That's interesting to think about.
Okay here's an example. Tell me if I'm right...

\[2y^7 \rightarrow 14y^6y'\] and \[2x^3 \rightarrow 6x^2\]

yes

I am sorry, but I am completely lost.

***I am sorry, but I am completely lost.***
do you know how to use the chain rule?

Yes, but the y and x dilemma was something never explained to me.

Oh! I think I'm starting to understand now!

yes

yes

yw