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

Let \[ y=x^{\frac{1}{2}}(x^2+3)^5. \] We differentiate using the product rule. Define \[ u=x^{\frac{1}{2}},\qquad v=(x^2+3)^5. \] Then, \[ \frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}, \] and by the chain rule, \[ \frac{dv}{dx}=5(x^2+3)^4\cdot\frac{d}{dx}(x^2+3)=5(x^2+3)^4\cdot2x=10x\,(x^2+3)^4. \] Applying the product rule: \[ \frac{dy}{dx}=u'v+uv'=\frac{1}{2}x^{-\frac{1}{2}}(x^2+3)^5+ x^{\frac{1}{2}}\cdot10x\,(x^2+3)^4. \] Simplify the second term: \[ x^{\frac{1}{2}}\cdot10x=10x^{\frac{3}{2}}, \] so that \[ \frac{dy}{dx}=\frac{1}{2}x^{-\frac{1}{2}}(x^2+3)^5+10x^{\frac{3}{2}}(x^2+3)^4. \] Factor out the common factors \(x^{-\frac{1}{2}}\) and \((x^2+3)^4\): \[ \frac{dy}{dx}=x^{-\frac{1}{2}}(x^2+3)^4\left[\frac{1}{2}(x^2+3)+10x^2\right]. \] Now, simplify the bracket: \[ \frac{1}{2}(x^2+3)+10x^2=\frac{1}{2}x^2+\frac{3}{2}+10x^2=\frac{1+20}{2}x^2+\frac{3}{2}=\frac{21x^2+3}{2}. \] Factor a common factor of \(3\): \[ \frac{21x^2+3}{2}=\frac{3(7x^2+1)}{2}. \] Thus, the derivative is: \[ \frac{dy}{dx}=\frac{3(x^2+3)^4}{2\,x^{\frac{1}{2}}}(7x^2+1). \]