To find [tex]\(\frac{d y}{d x}\)[/tex] for [tex]\(y = \sqrt{u}\)[/tex] and [tex]\(u = x^2 + 5\)[/tex], we can use the chain rule. The chain rule states that if a variable [tex]\(y\)[/tex] depends on a variable [tex]\(u\)[/tex], which in turn depends on a variable [tex]\(x\)[/tex], then the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] can be found using the formula:
[tex]\[
\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}
\][/tex]
Let's apply this step-by-step:
1. First, express [tex]\(y\)[/tex] in terms of [tex]\(u\)[/tex]:
[tex]\[
y = \sqrt{u}
\][/tex]
2. Compute [tex]\(\frac{d y}{d u}\)[/tex]:
[tex]\[
\frac{d y}{d u} = \frac{d}{d u} (\sqrt{u}) = \frac{1}{2\sqrt{u}}
\][/tex]
3. Now, express [tex]\(u\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[
u = x^2 + 5
\][/tex]
4. Compute [tex]\(\frac{d u}{d x}\)[/tex]:
[tex]\[
\frac{d u}{d x} = \frac{d}{d x} (x^2 + 5) = 2x
\][/tex]
5. Finally, combine these results using the chain rule:
[tex]\[
\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} = \frac{1}{2\sqrt{u}} \cdot 2x = \frac{2x}{2\sqrt{u}} = \frac{x}{\sqrt{u}}
\][/tex]
Since [tex]\(u = x^2 + 5\)[/tex], substitute [tex]\(u\)[/tex] back in terms of [tex]\(x\)[/tex]:
[tex]\[
\frac{d y}{d x} = \frac{x}{\sqrt{x^2 + 5}}
\][/tex]
Thus, the derivative [tex]\(\frac{d y}{d x}\)[/tex] is:
[tex]\[
\boxed{\frac{x}{\sqrt{x^2 + 5}}}
\][/tex]
