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If the first derivative f '(x) of F (x) is continuous, then ∫ XF' (x) DX= The one who answered XF (x) -∫ f (x) DX:
∫xf'(x)dx＝∫xdf(x)＝xf(x)－∫f(x)dx
-------
It cannot be simplified without other conditions
Let f (x) have a derivative at x = 2, then Lim f (2 + △ x) - f (2 - △ x) is equal to △ x → 02 △ x a.2f ′ (2) b.1/2f ′ (2) C. f ′ (2) D.4 f′(2) I'm a liberal arts student in senior two. I hope to use the formula of senior two
f'(2)=lim[f(2+△x)-f(2-△x)]/[(2+△x)-(2-△x)]=lim[f(2+△x)-f(2-△x)]/[2△x]
Lim f (2 + △ x) - f (2 - △ x) is equal to △ x → 0 2 △ X
Answer C. f '(2)
What is Lim [f (a + H) + F (A-H) - 2F (a)] / h ^ 2 equal to when h → 0 (let the derivative of F (x) be continuous from this neighborhood at x = a) In particular, I would like to ask whether f (a) is a constant in this formula and the derivative is 0. In addition, I would like to ask the whole application process of lobida's law
F (a) is a constant in this formula, and f '(a) is unknown. Lim [f (a + H) + F (A-H) - 2F (a)] / h ^ 2] = Lim [f' (a + H) + F '(A-H) (- 1)] / 2H = Lim [f' (a + H) - f '(a)] / 2H + Lim [f' (A-H) - f '(a)] / 2 (- H) = f' (a)
Let f (x) have a second derivative at x = a, and prove that LIM (f (a + x) + F (A-X) - 2F (a)) / x ^ 2 = f '' (a) when x tends to 0
It is known that f (x) has a second derivative at x = a, so the first derivative of F (x) is continuous in the critical domain of x = a
Derivative definition
Start proving
So the limit of the original formula is f''(a)
What is Lim [f (a + 2 h) - 2F (a + H) + F (a)] / h ^ 2 equal to when h → 0 (let the derivative of F (x) be continuous from this neighborhood at point x = a) Prove equal to f "(a)
lim（h→0）[f(a+2 h)-2f(a+h)+f(a)]/h^2
=lim（h→0）{f(a+2 h)-f(a+h)-[f(a+h)-f(a)]}/h^2
=lim（h→0）[f'(a+h)-f'(a)]/h
=f''(a)
Let f (x) be second-order differentiable in a field where x = 0, and Lim x→0(sin3x x3+f(x) X2) = 0, find f (0), f ′ (0), f ″ (0) and Lim x→0f(x)+3 x2．
Because:
lim
x→0
(
sin3x
x3
+
f(x)
x2
)＝
lim
x→0
sin3x+xf(x)
x3
＝
lim
x→0
sin3x
x
+f(x)
x2
＝0，
So:
lim
x→0
(
sin3x
x
+f(x))＝0．
Also: F (x) is second-order differentiable in a field where x = 0,
So: F (x), f '(x) is continuous at x = 0,
Thus: F (0) = - 3
from
lim
x→0
sin3x
x
+f(x)
x2
＝0，
Get:
lim
x→0
sin3x
x
−3+f(x)+3
x2
＝0，
It is also easy to know:
lim
x→0
3−
sin3x
x
x2
＝
lim
x→0
3x−sin3x
x3
＝
lim
x→0
3−3cos3x
3x2
=
lim
x→0
3sin3x
2x
＝
nine
two
，
Therefore:
lim
x→0
f(x)+3
x2
＝
nine
two
，
Thus: F '(0) =
lim
x→0
f(x)−f(0)
x−0
＝
lim
x→0
f(x)+3
x
＝
lim
x→0
x•
f(x)+3
x2
＝0 ×
nine
two
＝0，
F (x) is expanded by Taylor at x = 0
lim
x→0
f(x)+3
x2
＝
nine
two
Get:
lim
x→0
f(0)+f′(0)x+
one
2!
f″(0)x2+0(x2)+3
x2
＝
nine
two
，
Calculated:
one
two
f″(0)＝
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