Y = 2cos ^ 2x + 5sinx-4

Y = 2cos ^ 2x + 5sinx-4

Y = 2cos ^ 2x + 5sinx-4 = 2 (1-sin ^ 2x) + 5sinx-4 = - 2Sin ^ 2x + 5sinx-2 = - 2 (sin ^ 2x-5 / 2sinx) - 2 = - 2 (sinx-5 / 4) ^ 2 + 9 / 8 ∵ - 1 ≤ SiNx ≤ 1, so - 9 ≤ y ≤ 1, so SiNx = 1, ymax = 1

Finding the derivative of y = ln / x + 1 / LNX

y = ln(1/x) + 1/lnx = -lnx + 1/lnx
y' = -1/x (-1/x)/ln²x
= -(1/x)(1+1/ln²x)

The range of the function y = 2cos & sup2; X + 5sinx-4 is

y=2cos²x+5sinx-4
=2(1-sin²x)+5sinx-4
=-2-2sin²x+5sinxd
=-2(sinx-5/4)²+9/8
ymax=1 sinx=1
ymin=-9 sinx=-1
Range [- 9,1]

How to find the n-th derivative of y = ln (x + 1)

One by one: y '= (x + 1) ^ (- 1) = [(- 1) ^ (1-1)] (1-1)! (x + 1) ^ (- 1) y' '= - 1 (x + 1) ^ (- 2) = [(- 1) ^ (2-1)] (2-1)! (x + 1) ^ (- 2) y' '= 2 (x + 1) ^ (- 3) = [(- 1) ^ (3-1)] (3-1)! (x + 1) ^ (- 3) Y4 order derivation = - 6 (x + 1) ^ (- 4) = [(- 1) ^ (4-1)] (4-1)! (x + 1) ^

The more detailed the steps, the better

The square of COS = 2 of 1-sin

Find the n-th derivative of y = ln (2-x / 3 + x),

The range of the function y = 2cos (x − π 3), X ∈ [π 6], is______ ．

∵ π 6 ≤ x ≤ 2 π 3, ∵ - π 6 ≤ X - π 3 ≤ π 3, ∵ 12 ≤ cos (x - π 3) ≤ 1, ∵ 1 ≤ 2cos (x - π 3) ≤ 2, ∵ the range of function is [1,2], so the answer is [1,2]

Find the n-order derivative of y = ln (1 + x) and give the specific process,

y'=1/(1+x)=(1+x)^(-1)
y''=-1*(1+x)^(-2)
y'''=-1*(-2)*(1+x)^(-3)=2*(1+x)^(-3)
y''''=2*(-3)*(1+x)^(-4)=-6*(1+x)^(-4)
So y ^ (n) = (- 1) ^ (n + 1) * (n-1)! * (1 + x) ^ (- n)

What is the range of the function y = (3 + SiN x) / (4 + 2cos x)?

From y = (3 + SiN x) / (4 + 2cos x), sinx-2ycosx = 4y-3, so √ (4Y & sup2; + 1) sin (x + $) = 4y-3, so sin (x + $) = (4y-3) / √ (4Y & sup2; + 1) so | (4y-3) / √ (4Y & sup2; + 1) | ≤ 1, y ∈ [3 - (2 √ 3) / 3,3 + (2 √ 3) / 3]

Find the derivative of F (x) = ln / X /?
Finding the derivative of F (x) = ln / X /
The derivative of F (x) = a ^ - x
The extremum of F (x) = (x-1) ^ 4
Please write down the derivation process of the above three questions,

(1) When x > 0, LN | x | = LNX (LN | x |) '= (LNX)' = 1 / X; when x > 0, LN | x | = LNX (LN | x |) '= (LNX)' = 1 / X