1. We are given that $\lim_{h\to 0} \frac{1}{h}f(1+h)=5$ and $f(x)$ is differentiable at $x=1$.

2. Since $f$ is differentiable at $x=1$, it is also continuous at $x=1$. So as $h\to 0$, $f(1+h) \to f(1).$

3. Now analyze the limit $\lim_{h\to 0} \frac{f(1+h)}{h}.$ For this limit to be finite, we must have $f(1)=0,$ because if $f(1)\neq 0$, then near $h=0$, $f(1+h)\approx f(1)$ and so $\frac{f(1+h)}{h}$ would blow up.

4. Using $f(1)=0$, rewrite the given limit as $\lim_{h\to 0} \frac{f(1+h)-f(1)}{h}=5.$

5. But this is exactly the definition of the derivative at $x=1$: $f'(1)=\lim_{h\to 0} \frac{f(1+h)-f(1)}{h}.$

6. Therefore, $f'(1)=5.$

7. Checking options:

   • A: $3$ ❌
   • B: $4$ ❌
   • C: $5$ ✅
   • D: $6$ ❌

So the correct answer is C.