Using the Product and Quotient Rule

1. Differentiating \(y = (3 + x^2)(2+\ln x)\) requires the use of the product rule:
   Let \(y = u\times v\), where \(u = 3 + x^2\,\) and \(v = 2+ \ln x\)
   Differentiating \(u\) and \(v\) with respect to \(x\) gives: \[\frac{\mathrm{d}u}{\mathrm{d}x} = 2x \qquad \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{x} \]
   Substituting into the product rule gives:
   \begin{eqnarray*}
   \frac{\mathrm{d}(uv)}{\mathrm{d}x} &=& u \frac{\mathrm{d}v}{\mathrm{d}x} +v\frac{\mathrm{d}u}{\mathrm{d}x}\\
   &=& (3 + x^2) \times \frac{1}{x} + (2+ \ln x) \times 2x\\
   &=& \displaystyle\frac{(3 + x^2)}{x} + 2x(2 + \ln x) \\
   &=& \displaystyle\frac{3}{x} + x + 4x + 2x \ln x \\
   &=& \displaystyle\frac{3}{x} + 5x + 2x \ln x
   \end{eqnarray*}
2. Differentating \( \displaystyle y = \frac{x^2}{\sqrt{x-1}} \) (with respect to \(x\)) requires the use of the quotient rule:
   Let \(\displaystyle y = \frac{u}{v}\,\), where \(u = x^2\,\) and \(v = \sqrt{x-1}\,\).
   Differentiating \(u\) and \(v\) with respect to \(x\) gives:
   \[\frac{\mathrm{d}u}{\mathrm{d}x} = 2x \qquad \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2\sqrt{x-1}}\]
   Substituting into the quotient rule gives:
   \begin{eqnarray*}
   \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{u}{v}\right) & = & \frac{v\displaystyle\frac{\mathrm{d}u}{\mathrm{d}x} - u\displaystyle\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\\\\
   \frac{\mathrm{d}y}{\mathrm{d}x} &=& \frac{\sqrt{x-1} \times 2x - x^2 \times \displaystyle\frac{1}{2 \sqrt{x-1}}}{(\sqrt{x-1})^2}\\
   &=& \frac{2x\sqrt{x-1} - x^2\times \displaystyle\frac{1}{2\sqrt{x-1}}}{x-1}\\
   &=& \displaystyle\frac{\displaystyle\frac{2x\sqrt{x-1} \times \sqrt{x-1} - x^2\times \frac{1}{2}}{\sqrt{x-1}}}{x-1}\\\\
   &=& \displaystyle\frac{2x(x-1) - \frac{1}{2}x^2}{(x-1)(\sqrt{x-1})}\\\\
   &=& \displaystyle\frac{2x^2 - 2x - \frac{1}{2}x^2}{(x-1)(\sqrt{x-1})}\\\\
   &=& \displaystyle\frac{\frac{3}{2}x^2 - 2x}{(x-1)^{\frac{3}{2}}
   }
   \end{eqnarray*}