Maximum value and minimum value of y = sin2x / (SiNx + cosx + 1)

Maximum value and minimum value of y = sin2x / (SiNx + cosx + 1)

(sinx+cosx) ²= sin ² x+cos ² x+2sinxcosx
=1 + sin2x < = 1 + 1 = 2, if and only if sin2x = 1, i.e. x = 45 °, the equal sign is valid
So SiNx + Cox < = root 2
Y = SiNx + cosx + 1 < = 1 + radical 2
So the maximum value of the function y = SiNx + cosx + 1 is 1 + radical 2

If 0 ≤ x ≤ π, find the maximum and minimum values of the function y = sin2x + SiNx cosx

Since 0 ≤ x ≤ π, - pi / 4 ≤ x-pi / 4 ≤ 3 * pi / 4. Then - (root 2) / 2 ≤ SiNx (x-pi / 4) ≤ (root 2) / 2Y = sin2x + SiNx cosx = 1 - (SiNx cosx) ^ 2 + (SiNx cosx) = - [(SiNx cosx) - 1 / 2] ^ 2 + 5 / 4 = - [sin (x-pi / 4) - 1 / 2] ^ 2 + 5 / 4 twice the root, so when SiNx (x-pi / 4) = -

What is the derivative of Ln (1 + 1 / x)? What is the derivative of Ln (1 + 1 / x)?

See figure

Find the derivative of Ln (1-x)?

Let 1-x = a
Then (LNA) '= 1 / A
Original formula '= (LNA)' a '= 1 / (1-x) * (- 1) = 1 / (x-1)

What about LNA + LNB? lna-lnb

lna+lnb=ln(ab)
lna-lnb=ln(a/b)

If a + B = 1 L, LNA + LNB =?

lna+lnb = ln(ab)
The value of AB cannot be determined by a + B = 1
So it can't be solved
Unless there are other conditions, the value of AB can be determined