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How to find the n-th derivative of the 1 / x power of E?
First find the first order, second order and third order, and then find the law
What is the derivative of e to the X / 2 power
The X / 2 power of E is multiplied by 1 / 2
What is the derivative of the - 1 / 2 power of (3 + 2x-x Square)
y=(3+2x-x^2)^(-1/2)
y'=(-1/2)[(3+2x-x^2)^(-3/2)]*(3+2x-x^2)'
=(-1/2)[(3+2x-x^2)^(-3/2)]*(2-2x)
=(x-1)(3+2x-x^2)^(-3/2)
It is known that SiNx + cosx = 1 / 5, (- π / 2)
sinx+cosx=1/5
Square on both sides:
1+2sinxcosx=1/25
sinxcosx=-12/25
(3sin^2*x/2-2cosx/2sinx/2+cos^2*x/2）/tanx+（1/tanx）
=(2sin ² x/2-sinx+1)/1/sinxcosx
=(1-cosx-sinx+1)*(-12/25)
=(2-1/5)*(-12/25)
=-108/125
The vectors a = (cosx, SiNx), B = (2cosx / 2, - 2sinx / 2) are known, and X belongs to (- π / 9,2 π / 9] It is known that a · B = 2cos3x / 2 belongs to [1,2] |A-b| = 5-4cos3x / 2 under the root sign belongs to [1, root 3] Find the minimum value of FX = a · B - | A-B |
f(x)=a·b-|a-b|
=2cos3x/2-√(5-4cos3x/2)
In yuan, let √ (5-4cos3x / 2) = t
Then cos3x / 2 = (5-T) ²)/ 4 and t belongs to [1, √ 3]
∴f(x)=-0.5t ²- t+2.5
According to the quadratic function, the minimum value is 1 - √ 3
Verification: sin2x / [SiNx + (cosx-1)] [SiNx - (cosx-1)] = (1 + cosx) / SiNx
Original formula = 2sinxcosx / sin ^ 2x - (cosx-1) ^ 2 = 2sinxcosx / sin ^ 2x cos ^ 2x + 2cosx-1 = 2sinxcosx / - 2cos ^ 2x + 2cosx = SiNx / 1-cosx = 1 + cosx / SiNx
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