Suppose $f:\mathbb{R} \to \mathbb{R}$ is a function given by
$$f(x) =\begin{cases} 1 & \mbox{if} \ x=1 \\ e^{(x^{10}-1)}+(x-1)^2\sin\frac1{x-1} & \mbox{if} \ x\neq 1\end{cases}$$

(a) Find $f'(1)$

(b) Evaluate $\displaystyle \lim_{u\to\infty} \left[100u-u\sum_{k=1}^{100} f\left(1+\frac{k}{u}\right)\right]$.
### Part (a)
To find \( f'(1) \), we use the definition of the derivative:
\[ f'(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} \]

Given the function:
\[ f(x) = \begin{cases} 
1 & \text{if } x = 1 \\
e^{(x^{10} - 1)} + (x - 1)^2 \sin \left( \frac{1}{x - 1} \right) & \text{if } x \neq 1 
\end{cases} \]

We need to evaluate:
\[ f'(1) = \lim_{x \to 1} \frac{e^{(x^{10} - 1)} + (x - 1)^2 \sin \left( \frac{1}{x - 1} \right) - 1}{x - 1} \]

We can split this limit into two parts:
\[ f'(1) = \lim_{x \to 1} \left( \frac{e^{(x^{10} - 1)} - 1}{x - 1} + \frac{(x - 1)^2 \sin \left( \frac{1}{x - 1} \right)}{x - 1} \right) \]

First, consider the term:
\[ \lim_{x \to 1} \frac{e^{(x^{10} - 1)} - 1}{x - 1} \]

Using the substitution \( u = x^{10} - 1 \), as \( x \to 1 \), \( u \to 0 \). We have:
\[ \frac{e^u - 1}{u} \to 1 \text{ as } u \to 0 \]

Also, \( x^{10} - 1 \approx 10(x - 1) \) for \( x \) close to 1. Therefore:
\[ \lim_{x \to 1} \frac{e^{(x^{10} - 1)} - 1}{x - 1} = \lim_{x \to 1} \frac{e^{(x^{10} - 1)} - 1}{x^{10} - 1} \cdot \frac{x^{10} - 1}{x - 1} = 1 \cdot 10 = 10 \]

Next, consider the term:
\[ \lim_{x \to 1} \frac{(x - 1)^2 \sin \left( \frac{1}{x - 1} \right)}{x - 1} = \lim_{x \to 1} (x - 1) \sin \left( \frac{1}{x - 1} \right) \]

Using the substitution \( h = x - 1 \), as \( x \to 1 \), \( h \to 0 \). We have:
\[ \lim_{h \to 0} h \sin \left( \frac{1}{h} \right) \]

This limit is well-known and evaluates to 0 because \( \sin \left( \frac{1}{h} \right) \) is bounded between -1 and 1, and \( h \to 0 \).

Thus:
\[ \lim_{x \to 1} (x - 1) \sin \left( \frac{1}{x - 1} \right) = 0 \]

Combining both parts, we get:
\[ f'(1) = 10 + 0 = 10 \]

### Part (b)
We need to evaluate:
\[ \lim_{u \to \infty} \left[ 100u - u \sum_{k=1}^{100} f \left( 1 + \frac{k}{u} \right) \right] \]

First, rewrite the expression:
\[ \lim_{u \to \infty} \left[ 100u - u \sum_{k=1}^{100} f \left( 1 + \frac{k}{u} \right) \right] = \lim_{u \to \infty} \sum_{k=1}^{100} u \left[ 1 - f \left( 1 + \frac{k}{u} \right) \right] \]

We can interchange the limit and the sum:
\[ = \sum_{k=1}^{100} \lim_{u \to \infty} u \left[ 1 - f \left( 1 + \frac{k}{u} \right) \right] \]

Using the definition of the derivative:
\[ \lim_{u \to \infty} u \left[ 1 - f \left( 1 + \frac{k}{u} \right) \right] = k \lim_{h \to 0} \frac{1 - f(1 + h)}{h} = -k f'(1) \]

Thus:
\[ \sum_{k=1}^{100} k (-f'(1)) = -f'(1) \sum_{k=1}^{100} k \]

The sum of the first 100 natural numbers is:
\[ \sum_{k=1}^{100} k = \frac{100 \cdot 101}{2} = 5050 \]

Therefore:
\[ -f'(1) \sum_{k=1}^{100} k = -10 \cdot 5050 = -50500 \]

The final answer is \( \boxed{ -50500 } \)