[help] review the doubtful points in the process of proving two calculus inequalities in the whole book? In the process of Reading Mathematics Review Book (number three), we can't figure out why (as shown in the above screenshot) the individual steps of these two examples are always difficult to prove the inequality of calculus,

First of all, thank you for your reply! The second proof, when I look at it, is also understood as Sint ~ t, but I think it is a bit out of line and abrupt. I really don't understand how to think of this step. If I don't look at the answer, I can't even think of how to deal with it like this. I seem to be used to this when I'm trying to find the limit. I feel like it's just a collection

The first person to study the application of calculus in inequality
Who knows what relevant articles he has published that are more influential

The first one should be Leibniz. His Leibniz formula and integral mean value theorem can be used to prove integral inequality. He created calculus from the geometric point of view, which is on an equal footing with Newton
Downstairs, the differential mean value theorem in calculus has existed in the era of Lagrange. If the theory is not perfect, then the first one used for inequality should also be Lagrange, not Cauchy

Do you have an inequality proof using Taylor's mean value theorem

F (x) is second-order differentiable, and f (0) = 0, f (1) = 1, f '(0) = f' (1) = 0. It is proved that there should be x belonging to (0,1), such that | f '' (x) | > = 2. It is proved that by Taylor expansion: F (1 / 2) = f (0) + F '(0) * (1 / 2 - 0) + F "(P) * (1 / 2 - 0) ^ 2 (P belongs to (0,1 / 2)) f (1 / 2) = f (1) + F' (1) * (1 / 2 - 1) + F" (q) *

A calculus on the mean value theorem that part of the proof problem ~ actually quite simple ~ please~
It is known that the function f (x) is continuous on [0,1], differentiable in (0,1), and f (1) = 0. It is proved that there is a point C in (0,1), such that f '(c) = - f (c) / C
It shouldn't be difficult, but I'm incompetent Please help me write the process ~ thank you in advance^^~

Certification:
Let f (x) = XF (x)
Then the function f (x) is continuous on [0,1] and differentiable in (0,1), and f (1) = f (0) = 0
So there must be a point C such that f '(c) = 0, that is [XF (x)]' = 0, CF '(c) + F (c) = 0, f' (c) = - f (c) / C

X tends to zero, (x-sinx) / (x + SiNx)

1 + cosx / 1 + SiNx = 1 / 2 find the equation x = 2K π - 2arctan2 and x = 2K π process
The answer is 2K π - 2arctan2 and 2K π

^The square primitive equation can be reduced to: (sin ^ (x / 2) + cos ^ (x / 2) + 2Sin (x / 2) * cos (x / 2)) / 2cos ^ (x / 2) = 1 / 2, that is: (sin (x / 2) + cos (x / 2)) ^ / cos ^ (x / 2) = 1 (Tan (x / 2) + 1) ^ = 1, and the solution is Tan (x / 2) = 0, or tan (x / 2) = - 2, that is: x = 2K π or x = 2K π - 2arctan2

SiNx = 0, x = k π if cosx = 1, x =?

2kπ

Similar to sin (x + 2K π) = SiNx, cosx =?

Similar to sin (x + 2K π) = SiNx, cosx = cos (x + 2K π)

Let y ^ (98) = x [SiNx (LNX) - cos (LNX)], find y ^ (100)

Do you want to know which ^ sign represents the square?
If so, I really can't think of any good way
y^100 = y^98 * (y^98)^(1/49).
therefore
y^100 = (x[sinx(lnx)-cos(lnx)]) ^ (50/49)

Derivative f-prime (x) of function f (x) = SiNx + LNX

cosx+1/x

The first person to study the application of calculus in inequality Who knows what relevant articles he has published that are more influential