Finding the tangent line to a curve Find an equation for the tangent line to the curve 

$$x\sin(xy-y^2)=x^2-1$$ 

through the point $(1,1)$.
 A: Hint: You need to find the equation of the line
$$  \frac{y-y_0}{x-x_0}=m,  $$
where $m$ is the slope at the point $(x_0,y_0)$ which is given by
$$ m = \frac{dy}{dx}\Big|_{(x,y)=(x_0,y_0)}. $$
A: First of all make sure that your equation defines $y$ with respect to $x$. This allows you to know that the equation defines $y$ with respect to $x$ implicitly and so you can do the implicit differentiation. After that use the following way to find the certain slope $y'|_{x=1}$. If $F(x,y)=0$ defines $y$ with respect to $x$ implicitly, then $$y'=\frac{-F_x}{F_y}$$
A: When the curve is given by $y = f(x)$ then the slope of the tangent is $\frac{dy}{dx}$,then the equation of the tangent line can be found by the equation.$$y=\frac{dy}{dx}x+c$$the value of $c$ can be found with the help of the given point $(1,1)$. given:$$x\sin(xy-y^2)=x^2-1$$$$y^2-xy+arcsin\left(\frac{x^2-1}{x}\right)=0$$solve this quadratic equation in $y$ to get$$y=\frac{x+\sqrt{x^2-4arcsin\left(\frac{x^2-1}{x}\right)}}{2}$$the other root cancels as it does not satisfy $x=1,y=1$.find $\frac{dy}{dx}_{(1,1)}$$$\frac{dy}{dx}=\frac{1}{2}+\frac{2x-4\left(1+\frac{1}{x^2}\right)\left(\frac{1}{\sqrt{1-\frac{x^2-1}{x}}}\right)}{4\sqrt{x^2-4arcsin\left(\frac{x^2-1}{x}\right)}}$$$$\frac{dy}{dx}_{(1,1)}=-1$$therefore the equation of the tangent is$$y=2-x$$
