(a) Find, from first principles, the derivative of \(f(x) = (2x + 3)^{2}\).
(b) Evaluate : \(\int_{1} ^{2} \frac{(x + 1)(x^{2} - 2x + 2)}{x^{2}} \mathrm {d} x\)
(a) The formula for the first principles method of calculating derivatives is
\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}\)
\(f(x) = (2x + 3)^{2} = 4x^{2} + 12x + 9\)
\(f(x + \Delta x) = 4(x + \Delta x)^{2} + 12(x + \Delta x) + 9\)
= \(4x^{2} + 8x \Delta x + 4 (\Delta x)^{2} + 12x + 12\Delta x + 9\)
\(\lim \limits_{\Delta x \to 0} \frac{(4x^{2} + 8x \Delta x + 4(\Delta x)^{2} + 12x + 12\Delta x + 9) - (4x^{2} + 12x + 9)}{\Delta x}\)
= \(\lim \limits_{\Delta x \to 0} \frac{4(\Delta x)^{2} + (8x + 12)\Delta x}{\Delta x}\)
= \(\lim \limits_{\Delta x \to 0} 4 \Delta x + (8x + 12)\)
\(\frac{\mathrm d y}{\mathrm d x} = 8x + 12\)
(b) \(\int_{1} ^{2} \frac{(x + 1)(x^{2} - 2x + 2)}{x^{2}} \mathrm {d} x\)
= \(\int_{1} ^{2} \frac{x^{3} - 2x^{2} +2x + x^{2} - 2x + 2}{x^{2}} \mathrm {d} x\)
= \(\int_{1} ^{2} \frac{x^{3} - x^{2} + 2}{x^{2}} \mathrm {d} x\)
= \(\int_{1} ^{2} (x - 1 + 2x^{2}) \mathrm {d} x\)
= \([\frac{x^{2}}{2} - x - \frac{2}{x}]_{1} ^{2} \)
= \((\frac{2^{2}}{2} - 2 - \frac{2}{2}) - (\frac{1^{2}}{2} - 1 - \frac{2}{1})\)
= \((2 - 2 - 1) - (\frac{1}{2} - 1 - \frac{2}{1})\)
= \(-1 - (-\frac{5}{2}) = \frac{3}{2}\)
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