Implicit differentiation $e^{(x^2y)}=\cos y$

Question

I took the natural log of both sides, giving me $x^2y=\ln(\cos y)$.

Then I differentiated: $2xydx + (x^2)dy = (-\tan y)dy$

That turned into $\frac{dy}{dx} = \frac{2xy}{-\tan y-x^2}$

This was marked incorrect. I differentiated with another method and got the right answer. What did I do in this process that was illegal?

Answer 1

We have:

$e^{x^{2}y} = \cos(y)$

$x^{2}y = \ln(\cos(y))$

Implicitly differentiating:

$2xy + x^{2}y' = \frac{d}{dx}(\cos(y))\frac{1}{\cos(y)}$

$2xy + x^{2}y' = -\frac{y'\sin(y)}{\cos(y)}$

$2xy = y'(-\tan(y)-x^{2})$

$y' = -\frac{2xy}{\tan(y)+x^{2}}$

To be safe, you should use derivative notation instead of differentials because differentials aren't exactly "formal".

Answer 2

Your answer is correct. I would write it as $-\frac{2xy}{x^2+\tan y}$, but that's just a matter of personal preference.

I assume that the "correct answer" that was given was $\dfrac{dy}{dx}=-\displaystyle\frac{2xye^{x^2y}}{x^2e^{x^2y}+\sin y}$, because that's what Wolfram Alpha gives. However, this answer is exactly equivalent to yours. To see this, remember that $e^{x^2y}=\cos y$, so $$\frac{dy}{dx}=-\frac{2xy\cos y}{x^2\cos y+\sin y}=\frac{2xy\cos y}{x^2\cos y+\sin y}\cdot\frac{\sec y}{\sec y}=-\frac{2xy}{x^2+\frac{\sin y}{\cos y}}=\frac{2xy}{x^2+\tan y}$$ I would go talk to whoever marked it incorrect to see if the grade was a mistake. It could also be that your teacher/professor wanted you to do the problem a certain way, like not using $\ln$ and instead implicitly differentiating in the exponential. Either way, the result is correct.