# `y = x^(coshx) , (1, 1)` Find an equation of the tangent line to the graph of the function at the given point

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