Calc I: Eind the derivative (5) \( \frac{x^{3}-3}{\sqrt{2 x}} \)

We are given \[ f(x)=\frac{x^3-3}{\sqrt{2x}}. \] A good first step is to rewrite the expression by expressing \(\sqrt{2x}\) as \(\sqrt{2}\sqrt{x}\). This gives \[ f(x)=\frac{x^3-3}{\sqrt{2}\sqrt{x}}=\frac{1}{\sqrt{2}}(x^3-3)\,x^{-1/2}. \] We now differentiate using the product rule. Let \[ u(x)=x^3-3,\qquad v(x)=x^{-1/2}. \] Then, their derivatives are: \[ u'(x)=3x^2,\qquad v'(x)=-\frac{1}{2}x^{-3/2}. \] The derivative of \(f(x)\) is given by: \[ f'(x)=\frac{1}{\sqrt{2}}\left[u'(x)v(x)+u(x)v'(x)\right]=\frac{1}{\sqrt{2}}\left(3x^2\cdot x^{-1/2}+(x^3-3)\cdot\left(-\frac{1}{2}x^{-3/2}\right)\right). \] Simplify each term: \[ 3x^2\cdot x^{-1/2}=3x^{2-\frac{1}{2}}=3x^{\frac{3}{2}}, \] and \[ (x^3-3)\left(-\frac{1}{2}x^{-3/2}\right)=-\frac{1}{2}(x^3-3)x^{-3/2}. \] Thus, \[ f'(x)=\frac{1}{\sqrt{2}}\left(3x^{\frac{3}{2}}-\frac{1}{2}(x^3-3)x^{-3/2}\right). \] To combine the two terms into a single fraction, express \(3x^{\frac{3}{2}}\) with denominator \(2x^{\frac{3}{2}}\): \[ 3x^{\frac{3}{2}}=\frac{6x^{\frac{3}{2}}}{2}=\frac{6x^3}{2x^{\frac{3}{2}}}. \] Now, \[ f'(x)=\frac{1}{\sqrt{2}}\left(\frac{6x^3-(x^3-3)}{2x^{\frac{3}{2}}}\right)=\frac{1}{\sqrt{2}}\left(\frac{6x^3-x^3+3}{2x^{\frac{3}{2}}}\right). \] Simplify the numerator: \[ 6x^3-x^3+3=5x^3+3. \] Thus, the derivative is: \[ f'(x)=\frac{5x^3+3}{2\sqrt{2}\, x^{\frac{3}{2}}}. \]