find \( \frac{d y}{d x} \quad \begin{array}{r}\text { find } \\ \text { Tangent } \\ y=(x \\ \text { at }\end{array} \) 5) \( y=\sqrt{x}\left(x^{2}+3\right)^{5} \)

We are given \[ y = \sqrt{x}\,(x^2+3)^5 = x^{1/2}\,(x^2+3)^5. \] **Step 1. Express \(y\) as a product.** Let \[ u(x)=x^{1/2} \quad \text{and} \quad v(x)=(x^2+3)^5. \] **Step 2. Find \(u'(x)\) and \(v'(x)\).** Using the power rule, \[ u'(x)=\frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2}x^{-1/2}. \] For \(v(x)=(x^2+3)^5\), use the chain rule: \[ v'(x)=5\,(x^2+3)^4\cdot\frac{d}{dx}(x^2+3)=5\,(x^2+3)^4\cdot(2x)=10x\,(x^2+3)^4. \] **Step 3. Apply the product rule.** The product rule gives: \[ y'(x)=u'(x)v(x)+u(x)v'(x). \] Substitute the derivatives we found: \[ y'=\frac{1}{2}x^{-1/2}(x^2+3)^5+ x^{1/2}\cdot10x\,(x^2+3)^4. \] **Step 4. Simplify the expression.** Notice the second term can be written as: \[ x^{1/2}\cdot10x = 10x^{3/2}. \] Thus, \[ y' = \frac{1}{2}x^{-1/2}(x^2+3)^5 + 10x^{3/2}(x^2+3)^4. \] Factor out the common factors \(x^{-1/2}\) and \((x^2+3)^4\): \[ y' = x^{-1/2}(x^2+3)^4\left[\frac{1}{2}(x^2+3) + 10x^{3/2+1/2}\right]. \] Notice that \[ x^{3/2}=x^{-1/2}\cdot x^2 \quad \Longrightarrow \quad 10x^{3/2} = 10x^{-1/2}\cdot x^2. \] Thus, after factoring \(x^{-1/2}(x^2+3)^4\), we have: \[ y' = x^{-1/2}(x^2+3)^4\left[\frac{1}{2}(x^2+3) + 10x^2\right]. \] **Step 5. Write the final answer.** We can also combine the terms inside the brackets: \[ \frac{1}{2}(x^2+3) + 10x^2=\frac{1}{2}(x^2+3)+\frac{20x^2}{2}=\frac{x^2+3+20x^2}{2}=\frac{21x^2+3}{2}. \] Thus, one acceptable answer is \[ \boxed{y'=\frac{(21x^2+3)(x^2+3)^4}{2\sqrt{x}}.} \]