1. **State the problem:** We have differentiable functions $A(t), B(t), C(t), D(t)$ related by the equation $$AB = \log(C^2 + D^2 + 1).$$ We know at $t=9$ that $A(9) = 0$ and $\left. \frac{dA}{dt} \right|_{t=9} = 3$. We want to find $B(9)$ or determine if it does not exist (DNE). 2. **Analyze the given equation at $t=9$: ** At $t=9$, substitute $A(9) = 0$: $$0 \cdot B(9) = \log(C(9)^2 + D(9)^2 + 1)$$ which simplifies to $$0 = \log(C(9)^2 + D(9)^2 + 1).$$ 3. **Solve for $C(9)$ and $D(9)$: ** Since $\log(x) = 0$ implies $x=1$, we have $$C(9)^2 + D(9)^2 + 1 = 1 \implies C(9)^2 + D(9)^2 = 0.$$ Because squares are nonnegative, this means $$C(9) = 0 \quad \text{and} \quad D(9) = 0.$$ 4. **Differentiate both sides with respect to $t$: ** Using the product rule on the left side: $$\frac{d}{dt}[A(t)B(t)] = \frac{d}{dt} \left[ \log(C(t)^2 + D(t)^2 + 1) \right].$$ Left side: $$A'(t)B(t) + A(t)B'(t).$$ Right side: $$\frac{1}{C(t)^2 + D(t)^2 + 1} \cdot 2C(t)C'(t) + 2D(t)D'(t).$$ 5. **Evaluate at $t=9$: ** Substitute $A(9) = 0$, $C(9) = 0$, $D(9) = 0$, and $A'(9) = 3$: $$3 \cdot B(9) + 0 \cdot B'(9) = \frac{1}{1} \cdot (2 \cdot 0 \cdot C'(9) + 2 \cdot 0 \cdot D'(9)) = 0.$$ This simplifies to $$3 B(9) = 0 \implies B(9) = 0.$$ 6. **Conclusion:** We have found that $B(9) = 0$ based on the given information. **Final answer:** $$B(9) = 0.$$