Find $$ \frac{d y}{d x} $$ 5. $$ y=\sqrt{x}\left(x^{2}+3\right)^{5} $$

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Let 
$$
y = \sqrt{x}\,(x^2+3)^5.
$$
We can write this as a product of two functions:
$$
u(x)=\sqrt{x}=x^{\frac{1}{2}},\quad v(x)=(x^2+3)^5.
$$

**Step 1. Differentiate $$ u(x) $$:**

$$
u'(x)=\frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}.
$$

**Step 2. Differentiate $$ v(x) $$:**

Using the chain rule, let 
$$
w=x^2+3 \quad\text{so that}\quad v(x)=w^5.
$$
Then,
$$
v'(x)=5w^4\cdot \frac{dw}{dx}.
$$
Since
$$
\frac{dw}{dx}=2x,
$$
we have
$$
v'(x)=5(x^2+3)^4\cdot 2x=10x\,(x^2+3)^4.
$$

**Step 3. Apply the product rule:**

The product rule states
$$
\frac{d}{dx}(u\cdot v) = u'(x)v(x)+u(x)v'(x).
$$
Substituting, we obtain:
$$
\frac{dy}{dx}=\frac{1}{2\sqrt{x}}(x^2+3)^5 + \sqrt{x}\cdot 10x\,(x^2+3)^4.
$$

**Step 4. Factor common terms:**

Notice that both terms share a factor of $$(x^2+3)^4$$. Factoring it out, we get:
$$
\frac{dy}{dx}=(x^2+3)^4\left[\frac{x^2+3}{2\sqrt{x}}+10x\sqrt{x}\right].
$$

To combine the terms inside the bracket, rewrite $$10x\sqrt{x}$$ with a common denominator:
$$
10x\sqrt{x}=\frac{10x\sqrt{x}\cdot 2\sqrt{x}}{2\sqrt{x}}=\frac{20x^2}{2\sqrt{x}}.
$$
Thus, the expression inside the bracket becomes:
$$
\frac{x^2+3}{2\sqrt{x}}+\frac{20x^2}{2\sqrt{x}}=\frac{x^2+3+20x^2}{2\sqrt{x}}=\frac{21x^2+3}{2\sqrt{x}}.
$$

**Final Answer:**

$$
\frac{dy}{dx}=\frac{(x^2+3)^4(21x^2+3)}{2\sqrt{x}}.
$$
$$ \frac{dy}{dx} = \frac{(x^2 + 3)^4 (21x^2 + 3)}{2\sqrt{x}} $$

Find \( \frac{d y}{d x} \) 5. \( y=\sqrt{x}\left(x^{2}+3\right)^{5} \)

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