Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 35

Given,
y = (coshx - sinhx)^2 , (0, 1)
to find the tangent quation,
so first find the slope of the tangent and is as follows,
let y=f(x)
so we have to find the f'(x) to get the slope
so,
f'(x)= ((coshx - sinhx)^2)'
let u= (coshx - sinhx)
and (df)/dx = df/(du) * (du)/(dx)
so ,
f'(x) = d/du ( u^2) * d/dx (coshx -sinhx)
=2u* (d/dx(coshx) - d/dx(sinhx))
= 2u * (sinhx - coshx)
=2(coshx-sinhx)(sinhx-coshx)
so the slope of the line through the point (0,1) is
f'(x) = 2(coshx-sinhx)(sinhx-coshx)
f'(0) = 2(cosh 0-sinh 0 )(sinh 0-cosh 0)
= 2(1-0)(0-1)
= -2.
now the slope is -2 , so the equation of the tangent is ,
y-y_1= m (x-x1)
y-1=(-2)(x-0)
y-1=-2x
y=1-2x
so the tangent equation is y=1-2x