Question: Find $\frac{dy}{dx}$ implicity if $x=\tan(xy)$.
So far, I have used implicit differentiation on both sides of the equation and ended up with two different answers. The first one was a bit more work than the second one.
My first answer is $\frac{dy}{dx}=\frac{\cos^2{xy}(1-y\sec^2{xy})}{x}$.
My second answer is $1=y\sec^2{xy}+x\sec^2{xy}\frac{dy}{dx}$ therefore $\frac{dy}{dx}=\frac{1 - y\sec^2{xy}}{x\sec^2{xy}}$.
The wording of the question is what's really throwing me off. Which of these is the correct answer, or are they both wrong? Your help is always appreciated!
Consider $$F=\tan(xy)-x=0$$ So $$F'_x=y \sec ^2(x y)-1$$ $$ F'_y=x \sec ^2(x y)$$ and $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{\cos ^2(x y) \left(1-y \sec ^2(x y)\right)}{x}=\frac{\cos ^2(x y)-y}{x}$$ and this is exactly your second answer (before the last simplification).