is the question.
I started doing it then got stuck so I looked at the mark scheme:
Now, I don't understand the
$-sin(-2\pi)(x\frac{dy}{dx}+y)=0$ bit
$(x\frac{dy}{dx}+y)=0$
Solving $\frac{dy}{dx}=0$, substituting:
$(\sqrt{2\pi}\frac{dy}{dx}-\sqrt{2\pi})=0$
Wouldn't dy/dx here be equal to 1..?
The solution ignores the term after $-\sin(-2\pi)$ and doesn't bother to evaluate it, because immediately you can see that $-\sin(-2\pi) = 0$. Therefore, the equation reduces to the one for the $(0,0)$ case.
In general, the tangent line $y = -x$ has infinitely many such points with the given curve.
The equation after implicit differentiation is: $$e^{x+y}\biggl(1 + \frac{dy}{dx}\biggr) = - \sin (xy) \biggl(x \frac{dy}{dx} + y\biggr).$$ Let's consider what happens when we substitute $y = -x$: we get $$1 + \frac{dy}{dx} = - \sin (-x^2) \biggl(x \frac{dy}{dx} - x\biggr).$$ Now let's consider what happens when $x = \sqrt{2\pi}$: $$1 + \frac{dy}{dx} = - \sin (-2\pi) \biggl( \sqrt{2\pi} \frac{dy}{dx} - \sqrt{2\pi} \biggr).$$ But the RHS is ALWAYS zero, because $\sin(-2\pi) = 0$. Therefore, the above equation simplifies to $$1 + \frac{dy}{dx} = 0.$$ This is what is meant in the solution and in my earlier response. You have been solving the condition $$RHS = 0,$$ which is not what you actually want to do. You should be solving $$LHS = RHS.$$
