Given that the image of the function f (x) = SiNx + acosx is symmetric with respect to the straight line x = 5 π 3, then the value of the real number a is () A. −3B. −33C. 2D. 22

Given that the image of the function f (x) = SiNx + acosx is symmetric with respect to the straight line x = 5 π 3, then the value of the real number a is () A. −3B. −33C. 2D. 22

Y = SiNx + acosx = 1 + a2sin (x + φ), the maximum or minimum value is obtained at the axis of symmetry, ∩ sin5 π 3 + acos5 π 3 = ± 1 + A2, that is − 32 + 12a = ± 1 + A2, the solution is a = − 33; therefore, B is selected

(3x-2)（3x+2)-x(5x+1)>(2x+1)∧2

(3x-2)（3x+2)-x(5x+1)>(2x+1)²
9x²-4-5x²-x>4x²+4x+1
4x+1

The sum of polynomial - 8xy ^ 2 + 3x ^ 2Y and - 2x ^ 2Y + 5xy ^ 2 is--------
If a polynomial plus 5x ^ 2-4x-3 yields - x ^ 2-3x, then the polynomial is--------
-X + = - 5Y + 8y, then the polynomial in brackets is-------

-8xy^2+3x^2y+（-2x^2y+5xy^2）
=-3xy²+x²y
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4x-8 equals 3x

Is it to solve the equation
4x-8=3x
4x-3x=8
x=8

How to derive ln (x + 1) ^ 2?

ln(x+1)^2
Let u = x + 1, t = u ^ 2, y = LNT
[ln(x+1)^2]'
=(lnt)'*(t)'*(u)'
=[1/(x+2)^2]*2(x+1)*1
=2(x+1)/(x+2)^2
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Derivation of simple logarithmic function
What is the derivative of Ln (x + 1)? How to find the derivative of such a coincidence function

As t = x + 1 and y = LNT, where t is an intermediate variable
The derivation rule of compound function is as follows
The derivative of compound function is equal to the derivative of function to intermediate variable multiplied by the derivative of intermediate variable to independent variable
[f (t (x))] '= f' | t * t '| x (where the letter to the right of the vertical line "|" denotes the subscript)
y'=y'|t *t'|x
=1/t* 1
=1 / (x + 1)
If y = ln (x2 + x)
t=x2+x
y'=y'|t *t'|x
=1/t *( 2x+1)
=(2x+1)/(x2+1）
Another example is y = lncosx
t=cosx
y'=1/t *(-sinx)
=-sinx/cosx
=-TaNx (pay attention to reduction. The final result of any mathematical problem generally has the self-evident requirement of reduction)
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y'=1/t *(-sinx)

How to derive a logarithmic function?
Formula, derivative method

How to derive a logarithmic function?
The formula of the textbook is (Loga b) "= 1 / X · LNA
In particular, (LNX) "= 1 / X

Proof of derivative of logarithmic function
）f(x)=loga^x
f'(x)
=lim (loga^(x+Δx)-loga^x)/Δx
=lim loga^[(x+Δx)/x]/Δx
=lim loga^(1+Δx/x)/Δx
=lim ln(1+Δx/x)/(lna*Δx)
=lim x*ln(1+Δx/x)/(x*lna*Δx)
=lim (x/Δx)*ln(1+Δx/x)/(x*lna)
=lim ln[(1+Δx/x)^(x/Δx)]/(x*lna) ①
=lim lne/(x*lna) ②
=1/(x*lna)
① How can we deduce to (2)?

Using the basic limit: when X - > 0, (1 + x) ^ (1 / x) - > E

Derivative of y = (10 ^ x + 1) / (10 ^ x-1)

y'=[(10^x+1)'(10^x-1)-(10^x+1)(10^x-1)']/(10^x-1)^2
=[-2*10^x*ln10]/[10^x-1]^2

Finding the derivative of 1 / √ (x ^ 2 + 5) requires a detailed process

The derivative of the original formula = (√ (x ^ 2 + 5)) ^ (- 1 / 2)
=-（1/2）(√(x^2+5))^(-3/2)*2x
=-x/[(x^2+5)√(x^2+5)]