Find the tangent line to the curve I have a task that is described like this: Find the tangent line to the curve in the point (1,1)
I am also provided with the quation $x\sin(xy−y^2) = x^2−1$
I have followed some steps and ended up with $\sin(xy-y^2) + xy\cos(xy-y^2) = 2x$
Now i have a problem since i don't know how to proceed. I know that to find the tangent you often use y = mx + c, but i am struggeling when it comes to figuring this out. I hope that somebody is able to help.
 A: Yes you are right, you need $y=mx+c$ or $y=y'x+c$ to form the line tangent to a point $P$ on $y$
The equation below follows from your give after a few operations:
$$
xy-y^2=\sin^{-1}\left(
\frac{x^2-1}{x}
\right)
$$
by "completing the square":
$$
(y-x/2)^2-\frac{x^2}{4}=-\sin^{-1}\left(
\frac{x^2-1}{x}
\right)
$$
From here, find $m=y'$, and solve for $c$
A: Here is a graph of your relation:

Let’s use implicit differentiation for the slope:
$$\frac d{dx}[\sin(xy-y^2)+xy\cos(xy-y^2)=2x] \implies \cos(xy-y^2)\frac d{dx}[xy-y^2]+(xy’+y) \cos(xy-y^2) +xy\frac d{dx}[\cos(xy-y^2)]=2$$
Then, you should get the following after more chain and product rules. A bit of software was used to finish the messy derivatives:
$$-x y(x) (-2 y(x) y'(x) + x y'(x) + y(x)) \sin(x y(x) - y(x)^2) + x y'(x) \cos(x y(x) - y(x)^2) + (-2 y(x) y'(x) + x y'(x) + y(x)) \cos(x y(x) - y(x)^2) + y(x) \cos(x y(x) - y(x)^2)=2$$
Now set $x=1$ and solve for $y’(1)$ as the slope. For fun, let’s consider all possible slopes at $x=1$. Now remember that you should use point slope form with the final answer being:
$$y-y(1)=y’(1)(x-1)\implies y_{\text {tangent}}(x)=y’(1)(x-1)+y(1)$$
