Derivative of an implicit trigonometric function

Question

I was doing my homework which included finding the derivative of implicit function. One of them was

$$3\sin(xy)+ 4\cos(xy) = 5$$

On first attempt I did this as follows

$$\frac{d}{dx}(3\sin(xy)+ 4\cos(xy)) = \frac{d}{dx}(5)$$ $$[3\cos(xy)-4\sin(xy)][y+x\frac{dy}{dx}] = 0$$ $$\frac{dy}{dx}= \frac{-y}{x}$$

It matched the answer and I was happy. I don't know why but I wrote the original expression as this

$$\frac{3}{5}\sin(xy)+ \frac{4}{5}\cos(xy) = 1$$ $$\cos(\alpha)\sin(xy)+ \sin(\alpha)\cos(xy) = 1$$ $$\sin(xy +\alpha)=1 \quad \quad \textrm{where} \quad\alpha = \arccos(\frac{3}{5})$$

Again from step 2 of finding derivative,

$$[3\cos(xy)-4\sin(xy)][y+x\frac{dy}{dx}] = 0$$ $$[\frac{3}{5}\cos(xy)-\frac{4}{5}\sin(xy)][y+x\frac{dy}{dx}] = 0$$ $$[\cos(\alpha)\cos(xy)-\sin(\alpha)\sin(xy)][y+x\frac{dy}{dx}] = 0$$ $$[\cos(xy +\alpha)][y+x\frac{dy}{dx}] = 0$$

Since $\sin(xy +\alpha) =1 \implies \cos(xy+\alpha)=0$ what did in finding the derivate while going from step 2 to 3 was diving by zero which is not valid. So I cannot actually conclude that the derivative is $\frac{-y}{x}$. But then why is the answer right(wolfram alpha says this too)? Is there any other way of finding this derivate?

Thank you in advance.

Answer 1

From $\sin(xy + \alpha) = 1$, you can achieve that $xy + \alpha \in \{2k\pi + \frac{\pi}{2}\ |\ k \in \mathbb{Z}\}$ for all $x, y$. But $xy + \alpha$ is a continuous function of $x$ and thus, if its range is included in a discrete set, it must be constant. So $xy + \alpha = C$. Therefore, $y + xy' = 0$ and $y' = \frac{-y}{x}$.

Answer 2

Your problem is not suited for a direct application of the IFT. The function $f(x,y)=3\sin(xy) + 4\cos(xy) -5$ is smooth but if $f$ has a zero at $P(x_0,y_0)$ then as shown by your calculation $\partial_x f_{|P}= \partial_y f_{|P}=0$ so a priori you can not apply the IFT.

The function $f(x,y)=g(xy)=5 (\cos(xy+\alpha)-1)$ has isolated zeros when $xy+\alpha = 2\pi k$, $k\in {\Bbb Z}$ which leads to the wanted solution for the derivative but $g$ has zero derivative at the solution. If you replace the constant 5 on the RHS of your equation by any $c\in (-5,5)$ you do not run into this problem and the IFT works fine ok.

Answer 3

You didn't divide by zero. The reasoning is really "$AB = 0$ implies either $A=0$ or $B=0$". Then you can consider the cases $A=0$ and $A\neq 0$ separately. If $A\neq 0$, then it must be that $B=0$ (because then you can divide by the nonzero $A$) and you go on from there.