Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 36

The tangent line must go through the given point (x_0, y_0) and have the slope of y'(x_0). Thus the equation of the tangent line is
(y - y_0) = (x - x_0)*y'(x_0).
To find the derivative we need the chain rule and the derivative of sinh(x), which is cosh(x). Therefore y'(x) = (e^sinh(x))' =e^sinh(x)*cosh(x). For x = x_0 = 0 it is e^0*1 = 1.
So the equation of the tangent line is y - 1 = (x - 0)*1, or simply y = x + 1.
We have to check that y(x_0) = x_0. Yes, y(x_0) = y(0) = 1 and x_0 = 1.