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مشتق

مشتق. قضیه: اگر توابع f(x) و g(x) مشتق پذیر باشند آنگاه: [f(x) + g(x)]’ = f’(x) + g’(x) [f(x) - g(x)]’ = f’(x) - g’(x) [ Kf (x)]’ = Kf ’(x) [f(x)g(x)]’ = f’(x).g(x) + f(x).g’(x) [f(x)/g(x)]’ = f’(x)g(x) – f(x)g’(x) / [g(x)] 2 [g(x) ≠0] [ f n (x)]’ = nf n-1 (x)f’(x).

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مشتق

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  1. مشتق

  2. قضیه: • اگر توابع f(x) و g(x) مشتق پذیر باشند آنگاه: [f(x) + g(x)]’ = f’(x) + g’(x) [f(x) - g(x)]’ = f’(x) - g’(x) [Kf(x)]’ = Kf’(x) [f(x)g(x)]’ = f’(x).g(x) + f(x).g’(x) [f(x)/g(x)]’ = f’(x)g(x) – f(x)g’(x) / [g(x)]2 • [g(x) ≠0] [fn(x)]’ = nfn-1(x)f’(x)

  3. قضیه تابع چند جمله ای : • P(x) = a0xn + a2xn-1+ . . . an-1x + an • P’(x) = na0xn-1+ (n-1) a1xn-1. . . an-1

  4. قضیه قاعده زنجیری : • اگر توابع y=f(u) و u=g(x) مشتق پذیر باشند، آنگاه تابع مرکب y=(fog)(x) = f(g(x)) = f(u)

  5. y=(fog)(x) = f(g(x)) = f(u)

  6. f(x) = (x2+1)1/3+ 5x3/5 • f’(x) = 1/3(x2+1)-2/3.(2x) + 5(3/5)x-2/5

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