How would I solve the following question that is troubling me.
The question is
Find an equation for the tangent line to the graph of $x+\sin(y-2x)=1$ at point $(1,2)$
I did the following using the chain rule
$1+\cos(y-2x)(\frac{dy}{dx})(-2)$
then I simplified to $\frac{dy}{dx}=\frac{2}{cos(y-2x)}$ so I plugged in $x$ and $y $ and got $\cos(0)$ on the denominator which is $1$.
But I am unsure if I did that part correctly my final answer is $y-2=2(x-1)$
I think what is being suggested is that:
Differentiating $$x+\sin(y-2x)=1$$ by using the chain rule should result in:
$$1 + \cos(y - 2x)\left(\frac{dy}{dx} - 2\right) = 0$$
$$\frac{dy}{dx} = -\frac{-1 + 2\cos(y - 2x)}{\cos(y - 2x)},$$ then evaluate at point $(1,2)$