Let f (x) be differentiable at x = A and lim [f (a + 5H)] - f (a-5h)] / 2H = 1, then f '(a)=
The limit formula can be reduced to 5 / 2 (f '(a) + F' (a)) = 1, that is, 5F '(a) = 1, f' (a) = 1 / 5;
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- 1. Why can derivative be written as f '(a) = LIM (x → a) f (x) - f (a) / x-a Shouldn't it be written as f '(x) = LIM (△ x → 0) (x + Δ x) - f (x) / Δ x? I hope you can help me
- 2. It is proved that LIM (1-e ^ 1 / x) / (1 + e ^ 1 / x) does not exist when x tends to 0 LIM (1 / X of 1-e) / (1 / X of 1 + e) does not exist when x tends to zero
- 3. Let f (x) be differentiable on [0,1], and f (1) = 2 ∫ 0 ~ 1 / 2 XF (x) DX. It is proved that there exists ξ belonging to (0,1), such that f (ξ) + ξ f '(ξ) = 1
- 4. Given that f (x) is differentiable at x = 0, f (0) = 0, f '(0) = 2, what is the limit of F (sin3x) / X when x approaches 0
- 5. Let f (x) be differentiable at point X. = 0, and f (0) = 0 and f '(0) = 3, then the value of LIM (x →∞) [f (x) / x] ()
- 6. Let f (x) be differentiable and find Lim [f (x + △ x)] ^ 2 - [f (x)] ^ 2 △ x → 0 lim [f(x+△x)]^2-[f(x)]^2 △x→0 =lim{[f(x+△x)+f(x)]*[f(x+△x)-f(x)]}/△x Why is it equal to =2f(x)lim[f(x+△x)-f(x)]/△x Especially why is it equal to 2F (x) Please give specific reasons,
- 7. Let f (x) be differentiable in a neighborhood of x = 0, and lim f '(x) = 1, then f (x) has extremum in x = 0, and the detailed solution is obtained
- 8. X tends to 0) Lim [(x + 4) ^ 1 / 2-2] / sin3x
- 9. Lim x tends to 0 arcsin2x / sin3x
- 10. Let f (x) be differentiable at x0, then it is equal to Why Lim f (x + H) - f (X-H) / h = 2F '(x) Why 2F '(x), not 1 / 2 F' (x) F '(x) isn't that 2 in the denominator, and it becomes 1 / 2 after it is put forward?
- 11. Let f (x) be differentiable at x = 1 and f '(1) = 2, then [LIM (H → 0) f (1-h) - f (1)] / h is equal to
- 12. If the function f (x) is differentiable at x = a, then Lim h → 0 (f (a + 3H) - f (A-H)) △ 2H =?
- 13. Let f (x) be differentiable at x = 2 and f '(2) = 2, then Lim h → 0 [f (2-3H) - f (2)] / h =?
- 14. If f (x) is differentiable at point X and lim f (x-3h) - f (0) / h = 1, then f '(x) =? H-0
- 15. When Lim h tends to zero, (f (x0 + H) - f (x0-h)) / 2H = f '(x0) can't understand
- 16. If f ′ (x0) = - 2, then Lim [f (x0 + H) - f (x0-h)] / h=
- 17. If f ′ (x0) = - 3, then limh → 0f (x0 + H) - f (x0-h) H = () A. -3B. -6C. -9D. -12
- 18. If f ′ (x0) = - 3, then Lim [f (x0 + H) - f (x0-3h)] / h= process
- 19. The existence of partial derivative of binary function, why can we deduce the following first 1. X - > x0 Lim f (x, Y0) = y - > Y0 Lim f (x0, y) 2. F (x, y) is differentiable at P0. If a function of one variable is differentiable, the curve is smooth. Why is it not that a function of two variables is differentiable? What is the geometric meaning of differentiable function of two variables?
- 20. Is the existence of partial derivative the same as the existence of partial derivative? (function of two variables)