Derivative of $\ln(xy+1)=\sin(\pi x)$ at P(1,0) using implicit differentiation

Question

Firstly, I confirmed P(1,0) is on the curve by substitution.

Then I differentiated both sides giving me $\frac{x \frac {dy}{dx}+y}{xy+1}=\cos(\pi x)$

So $\frac {dy}{dx}=\frac {\cos(\pi x)(xy+1)-y}{x}$ But then evaluating $\frac{dy}{dx}$ at P(1,0) I get $-1$

The problem is that I know it should be $-\pi$ not $-1$

So basically where did I go wrong?

Answer 1

the derivative of $$(\sin(\pi x))'=\cos(\pi x)\pi$$

Answer 2

hint: $$\frac{d}{dx}(\sin(A x))= \frac{d}{dx}\sin(A x).\frac{d}{dx}(Ax)$$