I understand that we use the chain rule to differentiate the $y$ part of it. Like if it stood $xy$, I would use the product rule, that say that the derivative of $y$ is $dy/dx$. But what if it's just $y$ alone or with a constant in front off it?

Thank you!

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$\frac{\mathrm{d}}{\mathrm{d}x}(c\cdot f(x))=c\frac{\mathrm{d}}{\mathrm{d}x}f(x)$ if $c$ is a constant. –  Ｊ. Ｍ. Nov 7 '10 at 13:07
Thanks, if you put that in a reply I can check it off. –  Algific Nov 7 '10 at 13:36

If it's just $y$ alone then you don't need to use implicit differentiation, just differentiate the other side of the equation. If it's $cy$ then you can just divide both sides by $c$.

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Note that $\frac{d(f(x) \times y)}{dx} = \frac{d(f(x))}{dx} \times y + f(x) \times \frac{dy}{dx}$.

So for instance, if $f(x) = x$, we get $\frac{d(x \times y)}{dx} = \frac{d(x)}{dx} \times y + x \times \frac{dy}{dx} = 1 \times y + x \times \frac{dy}{dx} = y + x \times \frac{dy}{dx}$.

Similarly, if $f(x) = c$, where $c$ is a constant, then

$\frac{d(c \times y)}{dx} = \frac{d(c)}{dx} \times y + c \times \frac{dy}{dx} = 0 \times y + c \times \frac{dy}{dx} = c \times \frac{dy}{dx}$

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