^What do mathematical symbols mean

^What do mathematical symbols mean

5 ^ 2 is the second power of 5, and 6 ^ 3 is the third power of 6

What is the mathematical symbol for minutes

Hours: H
Minutes: minutes
Second: S

Let y = (e ^ x) - ln (cosx) and y = (e ^ x) - ln (cosx)

dy = [ e^x - 1/cosx *(-sinx) ] dx
= (e^x + tanx) dx

The answer to the indefinite integral is - ln | cosx | + C. why do I work out - 1 / cosx + C

∫tanx dx
=∫sinx/cosx dx
=-∫d(cosx)/cosx
=-ln|cosx|+C

How to find the integral of Ln (cosx)?
How to find the integral of Ln (cosx) from 0 to π / 2?
I understand the answer of 452192750. It's right. But I think it's like knowing the answer first and then putting it together. This method is too clever. I think it's hard for me to think of it. Is there any other way to think directly?

The answer is (- π / 2) LN2
(the following integrals are all definite integrals, and those not specified in the integral region are from 0 to π / 2)
1. Proof: ∫ ln (cosx) DX = ∫ ln (SiNx) DX. Let x = (π / 2) - t be substituted into the integral formula to get ∫ ln [cos ((π / 2) - t)] DT = ∫ ln (Sint) DT
2. Let the integral be I, then there is
2I+(π/2)ln2
=∫ln(cosx)dx+∫ln(sinx)dx+(π/2)ln2
=∫[ln(cosx)+ln(sinx)+ln2]dx
=∫ln(2cosxsinx)dx
=∫ln(sin2x)dx
3. Find out the relationship between ∫ ln (sin2x) DX and I
Let 2x = t, then
∫ln(sin2x)dx
=(1 / 2) ∫ ln (Sint) DT (integral region from 0 to π)
=(1 / 2) [∫ ln (Sint) DT + ∫ ln (Sint) DT] (the latter integral region is from π / 2 to π)
For ∫ ln (Sint) DT (the integral region is from π / 2 to π), substituting u = t - π / 2 into
It is equal to ∫ ln (COSU) Du, i.e. I
So there is ∫ ln (sin2x) DX = (1 / 2) [∫ ln (SiNx) DX + ∫ ln (cosx) DX] = I
Thus, according to the previous relation, there is 2I + (π / 2) LN2 = I, so I = (- π / 2) LN2

If the curve shown in the figure is the approximate picture of the function f (x) = X3 + bx2 + CX + D, then X12 + X22 equals___ ．

∵ f (x) = X3 + bx2 + CX + D, from the image, we know that - 1 + B-C + D = 0, 0 + 0 + 0 + D = 0, 8 + 4B + 2C + D = 0, ∵ d = 0, B = - 1, C = - 2 ∵ f ′ (x) = 3x2 + 2bx + C = 3x2-2x-2

Why Lim f (x0-h) - f (x0) / - H why f '(x0)

You may have misunderstandings about - H. I don't think you have any opinions about (f (x0 + H) - f (x0)) / h being derivative, but here h is a quantity that can be both positive and negative. If h takes positive and negative quantities, the result of this formula is different, then the limit does not exist according to the definition, and there is no derivative

If f '(0) = 2, find LIM (H → 0) f (x0-h) - f (x0) / 2H

lim(h→0)f(x0-h)-f(x0)/2h
=lim(h→0)(-1/2)*[f(x0)-f(x0-h)/h]
=(-1/2)*f'(0)
= -1

If Lim △ x → 0 f (x0 + △ x) - f (x0) / Δ x = k, then Lim △ x → 0 f (x0 + 2 △ x) - f (x0) / Δ x =?

Answer 2K
lim △x→0 f（X0+2△X）—f （X0）/△X
=(lim △x→0 f（X0+2△X）—f （X0）/2△X)*2
Do variable substitution DT = 2DX
Because DX - > 0, DT - > 0
=(lim dt→0 f（X0+dt）—f （X0）/dt)*2
Let DT = △ X
There is
2*lim △x→0 f（X0+△X）—f （X0）/△X =2k

If f (x) is differentiable and f (0) = f '(0) = √ 2, then LIM (H → 0) (f ^ 2 (H) - 2) / h=

Robida's law, first derivative, 2f (H) * f '(H),
h> The limit of > 0 is 2 * √ 2 * √ 2 = 4