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(D) To check the differentiability of $f(x)$ at $x=1$, we calculate the derivative using the definition: $f^{\prime}(1) = \lim_{h 	o 0} \frac{f(1+h)

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$f^{\prime}(1)$ exists and is $10$

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(D) To check the differentiability of $f(x)$ at $x=1$, we calculate the derivative using the definition: $f^{\prime}(1) = \lim_{h 	o 0} \frac{f(1+h) - f(1)}{h}$.Given $f(1) = 1$ and for $h 
eq 0$, $f(1+h) = e^{((1+h)^{10}-1)} + h^2 \sin(\frac{1}{h})$.Substituting these into the limit definition:$f^{\prime}(1) = \lim_{h 	o 0} \frac{e^{((1+h)^{10}-1)} + h^2 \sin(\frac{1}{h}) - 1}{h}$.Using the expansion $e^u = 1 + u + \frac{u^2}{2!} + \dots$, where $u = (1+h)^{10}-1 = 1 + 10h + \dots - 1 = 10h + O(h^2)$:$f^{\prime}(1) = \lim_{h 	o 0} \frac{1 + (10h + O(h^2)) + h^2 \sin(\frac{1}{h}) - 1}{h}$.$f^{\prime}(1) = \lim_{h 	o 0} \frac{10h + O(h^2) + h^2 \sin(\frac{1}{h})}{h}$.$f^{\prime}(1) = \lim_{h 	o 0} (10 + O(h) + h \sin(\frac{1}{h}))$.Since $\lim_{h 	o 0} h \sin(\frac{1}{h}) = 0$ by the Squeeze Theorem, we get $f^{\prime}(1) = 10 + 0 + 0 = 10$.

Suppose $f: R \rightarrow R$ is given by $f(x) = \begin{cases} 1, & \text{if } x=1 \\ e^{(x^{10}-1)} + (x-1)^2 \sin \frac{1}{x-1}, & \text{if } x \neq 1 \end{cases}$. Then:

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Reason $(R)$: If a function is differentiable at a point,then it is continuous at the point. But the converse is not true.

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