Answer
$$
-2
$$
Work Step by Step
We give that
\[
3=y \text { and } \sqrt{29-x^{2}-y^{2}}=z
\]
Thus:
\[
\begin{array}{l}
\sqrt{-x^{2}-(3)^{2}+29}=z \\
\sqrt{-x^{2}+20}=z
\end{array}
\]
Differentiating:
\[
\frac{d}{d x}(\sqrt{-x^{2}+20})=\frac{d z}{d x}
\]
\[
-\frac{x}{\sqrt{20-x^{2}+20}}=\frac{d z}{d x}
\]
Thus, at the point (4,3,2)
\[
\begin{array}{l}
\left.\frac{d z}{d x}\right|_{x=4}=-\frac{4}{\sqrt{20-4^{2}}} \\
\left.\frac{d z}{d x}\right|_{x=4}=-2
\end{array}
\]
