Answer to Question #269571 in Algebra for Tahmina

Question

Problem A.2

Let f(x), g(x) and h(x) be the functions from Problem A.1. Find the derivative λ′(x) of the following function with respect to x:

λ(x) = f(x) · g(x) + f(x) · h(x) − g(x) · h(x)

Answer 1

$\lambda(x) = f(x) · g(x) + f(x) · h(x) − g(x) · h(x) \\ Using \ the \ product \ rule \ to \ get \ derivative \ \lambda'(x) \\ \therefore \lambda'(x) = \\ f(x) · g'(x) + f'(x) · g(x) \\ + f(x) · h'(x) + f'(x) · h(x) \\ − g(x) · h'(x) + g'(x) · h(x) \\ From \ Problem\ A.1, \\ f(x) = x - 2 \\ g(x) = 4 - (x - 6)^2 + 2 \\ h(x) = x - 6 \\ \therefore \lambda(x) = \\ (x - 2) · (4 - (x - 6)^2 + 2) + (x - 2) · (x - 6) \\ - (4 - (x - 6)^2 + 2) · (x - 6) \\ f'(x) = \frac{d} {dx} (x-2) = 1 \\ g'(x) = \frac{d} {dx} (4-(x-6)^2+2) = -2(x-6) \\ h'(x) = \frac{d} {dx} (x-6) = 1\\ \lambda'(x) = \\ [(x-2) · (-2(x-6)) + 1 · (4-(x-6)^2 + 2)] \\ + [(x-2) · (1) + 1 · (x-6)] \\ - [(4-(x-6)^2 +2) · (1) + (-2 (x-6)) · (x-6)] \\ \lambda'(x) = \\ (x-2)(-2x+12) + (4-(x-6)^2 +2) \\ + (x-2) + (x-6) \\ - (4 - x^2 + 12x - 36 + 2 - 2x^2 + 24x - 72) \\ \lambda'(x) = \\ -2x^2 + 12x + 4x - 24 + 4 - x^2 + 12x - 36 \\ + 2 + x - 2 + x - 6 \\ - 4 + x^2 - 12x + 36 - 2 + 2x^2 - 24x + 72 \\ \lambda'(x) = -6x + 40 \\ \lambda'(x) = -2(3x - 20)$