To find the slope of the tangent line to the graph of $f(x) = \frac{e^{-4x}}{-2x^2}$ at $x = 1$, we need to compute $f'(x)$, the derivative of $f(x)$, and evaluate it at $x = 1$.

Step 1: Rewrite the function

The function is: $f(x) = \frac{e^{-4x}}{-2x^2}.$

Step 2: Apply the quotient rule

The quotient rule states: $\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}.$ Here:

• $u(x) = e^{-4x}$, so $u'(x) = -4e^{-4x}$ (by the chain rule).
• $v(x) = -2x^2$, so $v'(x) = -4x$ (using the power rule).

Substituting these into the quotient rule: $f'(x) = \frac{(-4e^{-4x})(-2x^2) - (e^{-4x})(-4x)}{(-2x^2)^2}.$

Step 3: Simplify the numerator

Expand the numerator: $(-4e^{-4x})(-2x^2) = 8x^2e^{-4x},$ $(e^{-4x})(-4x) = -4xe^{-4x}.$ Thus, the numerator becomes: $8x^2e^{-4x} - (-4xe^{-4x}) = 8x^2e^{-4x} + 4xe^{-4x}.$

Step 4: Simplify the denominator

The denominator is: $(-2x^2)^2 = 4x^4.$

Thus, the derivative is: $f'(x) = \frac{8x^2e^{-4x} + 4xe^{-4x}}{4x^4}.$

Step 5: Factor and simplify

Factor out $4xe^{-4x}$ from the numerator: $f'(x) = \frac{4xe^{-4x}(2x + 1)}{4x^4}.$ Cancel $4x$ from the numerator and denominator (note $x \neq 0$): $f'(x) = \frac{e^{-4x}(2x + 1)}{x^3}.$

Step 6: Evaluate at $x = 1$

Substitute $x = 1$ into $f'(x)$: $f'(1) = \frac{e^{-4(1)}(2(1) + 1)}{1^3}.$ Simplify: $f'(1) = \frac{e^{-4}(2 + 1)}{1} = 3e^{-4}.$

Final Answer:

The slope of the tangent line at $x = 1$ is: $\boxed{3e^{-4}}$

Would you like further clarification or details? 😊

---

Related Questions:

1. How does the quotient rule work, and why is it applied here?
2. What is the significance of simplifying expressions with negative exponents?
3. How do we interpret the slope of a tangent line geometrically?
4. Can the slope of a tangent line be zero? If so, under what conditions?
5. What is the difference between a derivative and a tangent line?

Tip: Always double-check your derivative calculations for complex functions, as small mistakes can propagate through the solution.