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

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`