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# Applying rules of differentiation

### 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*}