Gradient of a curve $y=\ln \sqrt{x+y}$

Question

Find the gradient of the curve $y=\ln \sqrt{x+y}$ at the point when its y-coordinate is 1.

My attempt,

I differentiated and I got $\frac{dy}{dx}=\frac{1}{2x+2y-1}$.

But I've problem in finding the $x$.

$y=\ln \sqrt{x+y}$

$1=\ln \sqrt{x+1}$

How to proceed then?

Answer 1

$$1=\ln \sqrt{x+1}\implies e^1=\sqrt{x+1}\implies e^2=x+1\implies x=e^2-1 $$

Answer 2

$$x+y=e^{2y}$$ Then $$2e^{2y}\frac{dy}{dx}=1+\frac{dy}{dx}$$ From which $$\frac{dy}{dx}=\frac{1}{2e^{2y}-1}$$

Answer 3

Why bother at all about x when

$$ \dfrac{dy}{dx} = \dfrac{1}{2 e^{ 2 y}-1 }? $$

Given answer is wrong, yours correct.

Answer 4

$y=\ln \sqrt{x+y}$

$\implies \sqrt{x+y}=e^y$

$\implies x+y=(e^y)^2$

$\implies x+y=e^{2y}$

$\implies 1+\frac{dy}{dx}=2e^{2y}\frac{dy}{dx}$

$\implies \frac{dy}{dx}=\frac{1}{2e^{2y}-1}$