`y = sinh(1-x^2) , (1, 0)` Find an equation of the tangent line to the graph of the function at the given point

First check that the given point `(x_0, y_0)` belongs to the given graph, i.e. that `y(x_0) = y_0.` Yes, `sinh(1-1^2) = 0.`

Then the equation of the tangent line is

`(y - y_0) = (x - x_0)*y'(x_0)`

(this line goes through `(x_0, y_0)` and has the required slope).

The derivative of `y` is `cosh(1 - x^2)*(-2x),` here we use the chain rule and the known derivative of `sinh.` Thus `y'(x_0) = y'(1) = -2*cosh(0) = -2.`

So the equation of the tangent line is  `y = (x - 1)*(-2) = -2x +2.`