MA100 Chapter Notes - Chapter 6: Jtc Corporation, Differentiable Function, Quotient Rule
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[k] = 0, k r d dx. [k f (x)] = k f (cid:48) (x), k r d dx. = [f (x) g (x)] = f (cid:48) (x) g(cid:48) (x) [xn] = nxn 1, n r d dx d dx (cid:20) f (x) g (x) (cid:21) [f (x) g (x)] = f (cid:48) (x) g (x) + f (x) g(cid:48) (x) f (cid:48) (x) g (x) f (x) g(cid:48) (x) Determine f(cid:48)(x) if f (x) = x2 + x + 1 x y = x4 + x + 1 x. Y(cid:48) = 3x2 + 0 x 2. = 3x2 1 x2 (6x + 1) (3x + 2) (cid:0)3x2 + x(cid:1) (3) (3x + 2)2. 18x2 + 15x + 2 9x2 3x (3x + 2)2. Given two or more di erentiable functions there are two methods of applying the chain rule. If y = x5 and x = 3t2 + t, nd y(cid:48)(t). Y(cid:48) = 5(3t2 + t)4(3(2t) + 1)
