1. **State the problem:** We are given functions $j(x)$ and $k(x)$ with their values and derivatives at $x=-1$, and a function $$f(x) = \frac{1 - k(x)}{j(x)^2 - 4}.$$ We need to find the derivative $f'(-1)$. 2. **Recall the formula:** To differentiate a quotient $$f(x) = \frac{u(x)}{v(x)},$$ use the quotient rule: $$$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}.$$ Here, $u(x) = 1 - k(x)$ and $v(x) = j(x)^2 - 4$. 3. **Find $u'(x)$ and $v'(x)$:** - $u'(x) = -k'(x)$ because derivative of 1 is 0 and derivative of $-k(x)$ is $-k'(x)$. - $v'(x) = 2j(x)j'(x)$ by chain rule since $v(x) = (j(x))^2 - 4$. 4. **Evaluate $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$ at $x=-1$ using the table:** - $j(-1) = 6$, $j'(-1) = -4$, $k(-1) = 5$, $k'(-1) = -6$. - $u(-1) = 1 - k(-1) = 1 - 5 = -4$. - $u'(-1) = -k'(-1) = -(-6) = 6$. - $v(-1) = j(-1)^2 - 4 = 6^2 - 4 = 36 - 4 = 32$. - $v'(-1) = 2 \times j(-1) \times j'(-1) = 2 \times 6 \times (-4) = -48$. 5. **Apply the quotient rule:** $$f'(-1) = \frac{u'(-1) v(-1) - u(-1) v'(-1)}{(v(-1))^2} = \frac{6 \times 32 - (-4) \times (-48)}{32^2}.$$ 6. **Simplify numerator:** $$6 \times 32 = 192,$$ $$(-4) \times (-48) = 192,$$ So numerator is $$192 - 192 = 0.$$ 7. **Simplify denominator:** $$32^2 = 1024.$$ 8. **Calculate final derivative:** $$f'(-1) = \frac{0}{1024} = 0.$$ **Final answer:** $$\boxed{0}.$$