May 16, 2017

Using $f ' \left(1\right) = {\lim}_{x \rightarrow 1} \frac{f \left(x\right) - f \left(1\right)}{x - 1}$, I get $f ' \left(1\right) = 10$

#### Explanation:

$f ' \left(1\right) = {\lim}_{x \rightarrow 1} \frac{f \left(x\right) - f \left(1\right)}{x - 1}$

$= {\lim}_{x \rightarrow 1} \frac{{e}^{{x}^{10} - 1} + {\left(x - 1\right)}^{2} \sin \left(\frac{1}{x - 1}\right) - 1}{x - 1}$

$= {\lim}_{x \rightarrow 1} \frac{{e}^{{x}^{10} - 1} - 1}{x - 1} + {\lim}_{x \rightarrow 1} \frac{{\left(x - 1\right)}^{2} \sin \left(\frac{1}{x - 1}\right)}{x - 1}$

Now, ${\lim}_{x \rightarrow 1} \frac{{e}^{{x}^{10} - 1} - 1}{x - 1}$ is $g ' \left(1\right)$ for $g \left(x\right) = {e}^{{x}^{10} - 1}$.

And, $g ' \left(x\right) = 10 {x}^{9} {e}^{{x}^{10} - 1}$, so $g ' \left(1\right) = 10$

The second limit,

${\lim}_{x \rightarrow 1} \frac{{\left(x - 1\right)}^{2} \sin \left(\frac{1}{x - 1}\right)}{x - 1} = {\lim}_{x \rightarrow 1} \left(\left(x - 1\right) \sin \left(\frac{1}{x - 1}\right)\right) = 0$ by the squeeze theorem.

Therefore,

$f ' \left(1\right) = 10$

Interestingly, the derivative exists at $1$, but is not continuous at $1$ because the limit as $x$ approaches $1$ of $f ' \left(x\right)$ fails to exist.

$f ' \left(x\right) = \left\{\begin{matrix}10 & \text{if" & x=1 \\ 10x^9e^(x^10-1)+2(x-1)sin(1/(x-1))-cos(1/(x-1)) & "if} & x \ne 1\end{matrix}\right.$