Question

How do you find the linearization of `e^sin(x)` at x=1?

Answer 1

The linearization of `f(x)=e^sinx` at `x=1` is

`f(x)~~f(1)+f'(1)*(x-1)`

Hence `f(1)=e^sin1~~2.32` and `f'(1)=e^(sin1)*cos1~~1.25`

Hence

`f(x)~~2.32+1.25*(x-1)`