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Find the derivative of function y = (x + 1) ^ 99
99x(x+1)^98
Find the derivative of function f (x) = x (x + 1)... (x + 99) (x + 100) by taking logarithm
lgf(x)=lgx+lg(x+1)+...lg(x+100)f'(x)/f(x)=1/x+1/(x+1)+..+1/(x+100)f'(x)=f(x)[1/x+1/(x+1)+...+1/(x+100)] =(x+1)...(x+100)+f(x)[1/(x+1)+..+1/(x+100)]f'(0)=100!+ 0=100!
F (x) is continuous on the closed interval and differentiable on the open interval. F (a) = f (b) = 1. It is proved that there are C and D belong to (a, b) so that (D / C) ^ (n-1) = f (c)+ c/n*f'(c)
It is proved that G (x) = x ^ NF (x), H (x) = x ^ n. from the properties of elementary functions, G (x), H (x) satisfy the condition of Lagrange mean value theorem on [a, b] ζ ∈ (a, b), so that G (b) - G (a) = g '（ ζ) (B-A) that is, f (b) B ^ n-f (a) a ^ n = B ^ n-a ^ n = [NF]（ ζ)ζ^ (n-1)+f'( ζ)ζ^ n] (B-A). (1) exists η ∈(a...
Let f (x) be continuous on [0, a], differentiable in (0, a), and f (a) = 0, it is proved that there is a point x belonging to (0, a), so that f (x) + X * f ` (x) = 0
Construct auxiliary function f (x) = XF (x)
F(0)=a F(a)=0
According to Rolle's theorem, there is a point x on (0, a) such that f '(x) = 0
That is, f (x) + XF '(x) = 0
Let f (x) be continuous on [0, a], differentiable in (0, a), and f (a) = 0. It is proved that there is at least one point C ∈ (0, a), so that f (c) + CF '(c) = 0 Followed by 3f (c) + CF '(c) = 0
Let g (x) = x ^ 3 F (x)
Then G (0) = g (a) = 0, according to the mean value theorem, there is C, 0
It is defined that the function on R satisfies f (- x) = 1 / F (x) > 0, and G (x) = f (x) + C (C is a constant) is a monotonic increasing function on [a, b]. It is proved that G (x) is monotonic on [- B, - A]
G (x) is a monotonically increasing function on [a, b]
That is, a < = x1g (x1) f (x) = g (x) - C
So f (x1) a < = X1 then - B < = - x2 < - X1 < = - A
f(-x)-=1/f(x)>0
So f (- x2) - f (- x1)
=1/f(x2)-1/f(x1)
=[f(x1)-f(x2)]/f(x1)f(x2)
1 / F (x) > 0, that is, f (x) > 0
So the denominator f (x1) f (x2) > 0
F (x1) so [f (x1) - f (x2)] / F (x1) f (x2) < 0
That is, when - B < = - x2 < - X1 < = - A, f (- x2) so f (x) increases
g(x)=f(x)+c
So g (- x2) - G (- x1) = f (- x2) - f (- x1) < 0
So monotonically increasing
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