implicit derivative of $\ln(xy)=\cos(x)-8y^2$

Question

Consider the equation

$$\ln(xy)=\cos(x)-8y^2$$

Now I thought the first step should be to raise everything to $e$ to get rid of the natural $\log$ but by doing that will only complicate the problem even more.

So instead I found the derivative of $\ln(xy)$ which is

$$\frac{dy}{dx} \ln(xy)= \frac{x+y^{'}}{xy}$$ Then finding the derivative of the 2nd term yeilds

$$\frac{dy}{dx}\cos(x)-8y^2=-\sin(x)-16y y^{'}$$

Then setting both of them equal to each other, like in the original equation gives

$$\frac{x+y^{'}}{xy}=-\sin(x)-16yy^{'}$$ But now I am lost on how to isolate $y^{'}$

Answer 1

$$ d\ln(xy)=d\cos x - 8d(y^2),\\ \frac{dx}{x}+\frac{dy}y=-(\sin x)dx-16ydy,\\ \frac1x+\frac{y'}y=-\sin x-16yy' $$

Now let's collect all terms with $y'$ on one side: $$ \frac{y'}y+16yy'=-\sin x-\frac 1x,\\ y'\frac{1+16y^2}{y}=-\frac{1+x\sin x}x,\\ y'=-\frac yx\frac{1+x\sin x}{1+16y^2} $$

Answer 2

The derivative of $\log (xy)$ is not quite right. You should have instead $$\frac{1}{xy}(xy)'=\frac{1}{xy}(y+xy').$$ Your derivative for RHS is correct, but the notation is bad.

Finally, set the two sides equal and solve for $y',$ which of course denotes the derivative of $y$ with regard to $x.$