Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 34

Given
y = x^(coshx) , (1, 1) to find the tangent line equation.
let
y=f(x)
so,
f(x) =x^(coshx)
so let's find f'(x) = (x^(coshx))'
on applying the exponent rule we get,
a^b = e^(b ln(a))
so ,
x^(coshx) = e^(coshx lnx)
so,
f'(x)= ( e^(coshx lnx))'
=d/dx ( e^(coshx lnx))
let u=coshx ln x
so ,
d/dx ( e^(coshx lnx)) = d/ (du) e^u * d/dx (coshx lnx)
= e^u * d/dx (coshx lnx)
=e^u * (sinhx lnx +coshx/x)
=e^(coshx ln x) * (sinhx lnx +coshx/x)
=> x^coshx * (sinhx lnx +coshx/x)
now let us find f'(x) value at (1,1) which is slope
f'(1) = x^cosh(1) (sinh(1) ln(1)+cosh(1)) = x^cosh(1) (0+cosh(1))
= x^cosh(1) (cosh(1))
now , the slope of the tangent line is x^cosh(1) (cosh(1))
we have the solope and the points so the equation of the tangent line is
y-y1 = slope(x-x1)
y-1=slope(x-1)
y= slope(x-1) +1
=x^cosh(1) (cosh(1)) (x-1)+1
but x^cosh(1) = e^(cosh(1)ln(1)) = e^0 =1
so,

=1 (cosh(1)) (x-1)+1
= xcoshx -coshx +1
so ,
y=xcoshx -coshx +1 is the tangent equation