The problem asks to find $f'(1)$ using the definition of the derivative for the function $f(x) = \frac{4}{x^2}$. The definition of the derivative is given by the following limit:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

To find $f'(1)$, we use this definition at $x = 1$:

$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$

Step-by-step solution:

1. Step 1: Compute $f(1+h)$ and $f(1)$:

   • $f(1) = \frac{4}{1^2} = 4$
   • $f(1+h) = \frac{4}{(1+h)^2}$

2. Step 2: Substitute these into the derivative definition: $f'(1) = \lim_{h \to 0} \frac{\frac{4}{(1+h)^2} - 4}{h}$

3. Step 3: Simplify the expression: $f'(1) = \lim_{h \to 0} \frac{\frac{4}{(1+h)^2} - \frac{4(1+h)^2}{(1+h)^2}}{h}$ $= \lim_{h \to 0} \frac{4\left( \frac{1 - (1+h)^2}{(1+h)^2} \right)}{h}$

4. Step 4: Expand and simplify the numerator: $= \lim_{h \to 0} \frac{4\left(1 - (1 + 2h + h^2)\right)}{h(1 + h)^2}$ $= \lim_{h \to 0} \frac{4\left(- 2h - h^2 \right)}{h(1 + h)^2}$ $= \lim_{h \to 0} \frac{-8h - 4h^2}{h(1 + h)^2}$

5. Step 5: Cancel the $h$ in the numerator and denominator: $= \lim_{h \to 0} \frac{-8 - 4h}{(1+h)^2}$

6. Step 6: Evaluate the limit as $h \to 0$: $f'(1) = \frac{-8}{1^2} = -8$

Thus, $f'(1) = -8$.

Would you like further details on any step? Here are some related questions to deepen your understanding:

1. How does the definition of the derivative relate to the slope of the tangent line?
2. What would happen if you tried to compute $f'(x)$ for other values of $x$?
3. How can we generalize this method to any rational function?
4. What if the function were $f(x) = \frac{1}{x^2}$ instead of $\frac{4}{x^2}$? How would the answer change?
5. How do the properties of limits help simplify complex derivative expressions?

Tip: When simplifying expressions involving limits, always check if common terms in the numerator and denominator can be factored and canceled before evaluating the limit.