(a) If \(y = (2x + 3)^{7} + \frac{x + 1}{2x - 1}\), find the value of \(\frac{\mathrm d y}{\mathrm d x}\) at x = -1. (b) Using the substitution, \(u = x + 2...
Question 1 Report
(a) If \(y = (2x + 3)^{7} + \frac{x + 1}{2x - 1}\), find the value of \(\frac{\mathrm d y}{\mathrm d x}\) at x = -1.
(b) Using the substitution, \(u = x + 2\), evaluate \(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x\).
Answer Details
(a) Differentiate \(y = (2x+3)^7 + \dfrac{x+1}{2x-1}\). For the first term use the chain rule; for the second use the quotient rule:
\[\frac{dy}{dx} = 7(2x+3)^6(2) + \frac{(2x-1)(1) - (x+1)(2)}{(2x-1)^2} = 14(2x+3)^6 + \frac{-3}{(2x-1)^2}.\]
At \(x = -1\): \(2x+3 = 1\) so \((2x+3)^6 = 1\), and \(2x-1 = -3\) so \((2x-1)^2 = 9\):
\[\frac{dy}{dx} = 14(1) - \frac{3}{9} = 14 - \frac{1}{3} = \frac{41}{3} = 13\tfrac{2}{3}.\]
(b) With \(u = x + 2\), we have \(x = u - 2\), \(x - 1 = u - 3\), \(dx = du\). Limits: \(x=1\Rightarrow u=3\), \(x=2\Rightarrow u=4\).
\[\int_1^2 \frac{x-1}{(x+2)^4}\,dx = \int_3^4 \frac{u-3}{u^4}\,du = \int_3^4 \left(u^{-3} - 3u^{-4}\right)du.\]
\[= \left[-\frac{1}{2u^2} + \frac{1}{u^3}\right]_3^4 = \left(-\frac{1}{32} + \frac{1}{64}\right) - \left(-\frac{1}{18} + \frac{1}{27}\right) = -\frac{1}{64} + \frac{1}{54} = \frac{5}{1728} \approx 0.0029.\]
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