# What is the local linearization of e^sin(x) near x=1?

Aug 4, 2017

$f \left(x\right) \approx 1.2534 x + 1.0664$

#### Explanation:

Let:

$f \left(x\right) = {e}^{\sin x}$

The linear approximation of a function $f \left(x\right)$ about $x = 1$ is given by the linear terms of the Taylor Series about the point $x = a$, ie the terms given by:

$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

Differentiating $f \left(x\right)$ using the chain rule, we have:

$f ' \left(x\right) = {e}^{\sin x} \left(\cos x\right)$

So with $x = 1$ we have:

$f \left(1\right) = {e}^{\sin 1}$
$\text{ } = 2.3197768 \ldots$

$f ' \left(1\right) = {e}^{\sin 1} \left(\cos 1\right)$
$\text{ } = 1.2533897 \ldots$

Hence, the linear approximation near $x = 1$ is given by:

$f \left(x\right) \approx 2.3198 + 1.2534 \left(x - 1\right)$
$\text{ } = 2.3198 + 1.2534 x - 1.2534$
$\text{ } = 1.2534 x + 1.0664$

Where we have rounded to 4dp.

Example:

Consider the case $x = 1.01$ which is near $x = 1$, Then:

$f \left(1.01\right) = {e}^{\sin \left(1.01\right)}$
$\text{ } = 2.332246 \ldots$

And the linear approximation gives us:

$f \left(1.01\right) \approx 1.2534 \left(1.01\right) + 1.0664$
$\text{ } = 1.265934 + 1.0664$
$\text{ } = 2.332334$

Which is correct within $3 \mathrm{dp}$