Question da527

Nov 19, 2016

I think there is an $=$ sign missing but....

Explanation:

Here you have an Implicit Derivation; in this case the function is kind of "hidden" so that you cannot really write it in the normal, explicit: $y = f \left(x\right)$.
Our idea is that $y$ represents a function of $x$ so that when you derive you must include the term $\frac{\mathrm{dy}}{\mathrm{dx}}$ to indicate the fact the $y$ is, in itsef, a function of $x$.
So for example if you have ${y}^{2}$ this derived will give you:
$2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

In your case I suspect that probably there should be a $=$ sign somewhere such as:
$8 x y + 16 \sin \left(y\right) = 0$
or other (you need the $=$ sign to indicate the functional dependence such as: "this depends from that");

but anyway, if you ONLY derive your expression you should get:

$8 y + 8 x \frac{\mathrm{dy}}{\mathrm{dx}} + 16 \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Now it ends here...on the other hand if you consider the original as:
$8 x y + 16 \sin \left(y\right) = 0$
you'll get:
$8 y + 8 x \frac{\mathrm{dy}}{\mathrm{dx}} + 16 \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
you can now collect $\frac{\mathrm{dy}}{\mathrm{dx}}$, rearrange and write:
(dy)/(dx)=-(cancel(8)y)/(cancel(8)(x+2cos(y)))=-(y)/((x+2cos(y))#