Let f (x) be continuous on [0,1], and (0,1) be derivable. F (0) = 0, f (1) = 1. It is proved that the existence of C belongs to (0,1) so that f (c) = 1-c Let f (x) be continuous on [0,1], and (0,1) be derivable. F (0) = 0, f (1) = 1. It is proved that the existence of C belongs to (0,1) so that f (c) = 1-c

Let f (x) be continuous on [0,1], and (0,1) be derivable. F (0) = 0, f (1) = 1. It is proved that the existence of C belongs to (0,1) so that f (c) = 1-c Let f (x) be continuous on [0,1], and (0,1) be derivable. F (0) = 0, f (1) = 1. It is proved that the existence of C belongs to (0,1) so that f (c) = 1-c

prove:
Let f (x) = f (x) + X-1
Because f (x) is continuous on [0,1] and differentiable at (0,1),
So f (x) is continuous on [0,1] and differentiable on (0,1),
also
F(0)=f(0)+0-1=-1<0
F(1)=f(1)+1-1=1>0
F (x) must have zero in [0,1], so there is f (c) + C-1 = 0, that is, f (c) = 1-c

Let f (x) be differentiable everywhere and f (0) = 0. It is proved that for any b > 0, there is C ∈ (0, b) so that f '(c) = f (b) / b

According to Lagrange mean value theorem:
existence α ∈(0,b),
Such that: [f (b) - f (0)] / (B-0) = f '（ α)= f(b)/b
Let C be equal to this α All right

Find the derivative of LNX ^ 2x + LNX ^ 2

y = ln(x^(2x)) + ln(x^2) = ln((x^2)^x) + ln(x^2) = (x+1)lnx^2
dy = ln(x^2)dx + (x+1)(1/x^2)2xdx = (2(x+1)/x) + lnx^2)dx
dy/dx = (2(x+1)/x) + lnx^2)
Note that when calculating the derivative, LNX ^ 2 cannot be transformed into 2lnx, because this transformation changes the definition field of the function. Obviously, the original function is a non-zero r as the definition field. If you finally get the result that LNX ^ 2 is written as 2lnx, then the derivative function is undefined in the negative number

F (x) = derivative of LNX ^ 2

f(x)=lnx^2
f'(x)=1/x^2 *(x^2)'
=2x/x^2
=2/x

Let f (x) = X^ λ Cos1 / X (x ≠ 0) = 0 (x = 0), and its derivative is continuous at x = 0 λ The answer is λ＞ 2. I really don't understand the answer,

f'(x)= λ x^( λ- 1)-sin1/x*(-1/x^2)
limf'(x)∞→0-= limf'(x)∞→0+

Let f (x) = x ^ a × Cos (1 \ x), X ≠ 0, f (x) = 0, x = 0, its derivative is continuous at x = 0, and the value range of a is? The answer to this question on the Internet is: using the definition of F '(0) = LIM (f (x) - f (0)) / (x-0) = limx ^ {A-1) cos (1 / x) = 0, when a > 1, if a ＜ 1 has no limit, it is non differentiable. Where it is not equal to 0, using the derivation rule of elementary function, there are f' (x) = ax ^ {A-1) cos (1 / x) - x ^ {A-2} sin (1 / x). When x tends to 0, there must be a > 2, then the above formula tends to 0. At this time, it is continuous, so when a > 2, the derivative is continuous I have a question: the above answer says that f '(x) = ax ^ {A-1) cos (1 / x) - x ^ {A-2} sin (1 / x). When x tends to 0, there must be a > 2 before the above formula tends to 0. If x = 0.000001 and A-2 = 0.001, doesn't x ^ {A-2} tend to 1? Isn't f' (x) formula tending to 0?

When A-2 = 0.001, x ^ (A-2) tends to 0, sin (1 / x) is bounded, and the product of the two tends to 0. The first term also tends to 0, and the difference between the two terms is 0