(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).
Explanation
(a) (y = (2x + 3)^{7} + frac{x – 1}{2x – 1})
(frac{mathrm d y}{mathrm d x} = 7(2x + 3)^{6}. 2 + frac{(2x – 1).1 – (x – 1).2}{(2x – 1)^{2}})
= (14(2x + 3)^{6} + frac{2x – 1 – 2x + 2}{(2x – 1)^{2}})
= (14(2x + 3)^{6} + frac{1}{(2x – 1)^{2}})
At x = -1, (frac{mathrm d y}{mathrm d x} = 14(-1)^{6} + frac{1}{(2(-1) – 1)^{2}})
= (frac{mathrm d y}{mathrm d x} = 14 + frac{1}{9} = 14frac{1}{9})
(b) (int_{1} ^{2} frac{x – 1}{(x + 2)^{4}} mathrm d x)
( u = x + 2 implies mathrm d u = mathrm d x)
If x = 1, u = 1 + 2 = 3; x = 2, u = 2 + 2 = 4.
(int_{1} ^{2} frac{x – 1}{(x + 2)^{4}} mathrm d x = int_{3} ^{4} frac{u – 3}{u^{4}})
= (int_{3} ^{4} (frac{u}{u^{4}} – frac{3}{u^{4}}) mathrm d u)
= (int_{3} ^{4} (u^{-3} – 3u^{-4}) mathrm d u)
= ([frac{u^{-2}}{-2} + u^{-3}]_{3} ^{4})
= ([frac{1}{u^{3}} – frac{1}{2u^{2}}]_{3} ^{4})
= ((frac{1}{4^{3}} – frac{1}{2(4^{2})}) – (frac{1}{3^{3}} – frac{1}{2(3^{2})}))
= (frac{1}{64} – frac{1}{32} – frac{1}{27} + frac{1}{18})
= (frac{5}{1728}).