Find the derivative of the function y = ln √ (x / 1 + x ^ 2)

Find the derivative of the function y = ln √ (x / 1 + x ^ 2)

y=(1/2)(ln(x)-ln(1+x^2)) y‘=(1/2)(1/x-2x/(1+x^2))=(1/2)((1-x^2)/(1+x^2)

Find the derivative of the function y = ln (LNX)

∵ function y = ln (LNX),
∴y′=1
lnx•(lnx)′=1
xlnx．
The derivative of the function y = ln (LNX) is 1
xlnx．

Find the derivative of function y = xlnx Can you tell me how to find this composite function?

F(x)=G(x) H(x)
F'(x)=G'(x)H(x) G(x)H'(x)
therefore
y=1*lnx x*1/x
=lnx 1
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Find n-th derivative of Y with known function y = xlnx

Y=XLnX
Y’=LnX+1 Y’’=1/X
Y(n)=(Y’’)(n-2)
=(1/X)(n-2)
=(-1)n/Xn-1
Y(n) = LnX+1 (n=1)
= (-1)n/Xn-1 (n>1)
Note: some of the above are superscripts. Those with brackets indicate n-order derivatives and those without brackets represent power exponent

The general expression of n-th derivative of function y = xlnx

First of all, y '= LNX + 1
Second order y '' = x ^ (- 1)
Third order y '' '= - x ^ (- 2)
Fourth order y (4) = x ^ (- 3)
If n is even, then y (n) = x ^ (- N + 1)
When it is an odd number, it is expressed as y (n) = - x ^ (- N + 1)

[derivative] given the function f (x) = ln (1 + x ^ 2) - 1 / 2x ^ 2 + m, the number of zeros of F (x) is discussed

The domain of F (x) is r
f'(x)=2x/(1+x^2)-x=x(1-x^2)/(1+x^2)
From F '(x) = 0, the extreme point x = - 1,0,1, so there are at most four zeros
F (- 1) = ln2-1 / 2 + m is the maximum
F (0) = m is the minimum
F (1) = ln2-1 / 2 + m is the maximum
While f (- ∞) = - ∞, f (+ ∞) = - ∞
Discussion M:
1) If M > 0, then f (- 1) = f (1) > 0, f (0) > 0, there are two zeros, respectively located at x1
2) If M = 0, then there are three zeros at X1 and x = 0
3) If 1 / 2-ln2

Given the function f (x) = xlnx, if the function g (x) = f (x) + x ^ 2 + ax + 2 has a zero point, find the minimum of the real number a If any value x > 0, f (x) / X ≤ x-kx ^ 2-1 holds, find the value range of real number K

Given the function f (x) = xlnx1, if the function g (x) = f (x) + x ^ 2 + ax + 2 has zero point, find the maximum value of real number a 2. If any value x is greater than 0, f (x) / X is less than or equal to x-kx ^ 2-1, it is tenable to find the value range of real number k (1) analytic: ∵ function f (x) = xlnx, Let f '(x) = LNX + 1 = 0 = = = > x = 1 / E, f' '(x) = 1 / x > 0 ? function f (x

Given the function f (x) = xlnx + 2x, find the derivative of y = f (x)

y'=(xlnx)'＋(2x)'
=(xlnx)'＋2
=(x)'lnx＋(x)(lnx)'＋2
=lnx＋1＋2
=lnx＋3

The general expression of n-order derivative of y = xlnx If the title, write the process method, thank you!

y'=lnx+1,
y"=1/x=x^(1-2)*(-1)^2,
The following orders are indicated by numbers in brackets,
y(3)=-1/x^2=x^(1-3)*(-1)^3=(3-2)!*x^(1-3)*(-1)^3,
y(4)=(4-2)!*x^(1-4)*(-1)^4,
y(5)=(5-2)!*x^(1-5)*(-1)^5
.
y(n)=(n-2)!*x^(1-n)*(-1)^n,(n∈N,n>=2）.
When n = 1, y '= 1 / x + 1,
n> When = 2,
y(n)=(n-2)!*x^(1-n)*(-1)^n,(n∈N,n>=2）.
The factorial of 0 is defined as 1 and! Is the factorial symbol

1. Find the monotone interval of the function f (x) = 2x'2-lnx (x > 0). 2: find the extremum of the function y = 2x + 8 / X

f(x)=2x'2-lnx ,f'(x)=4x-1/x =(4^2-1)/x
f'(x)=0 ,x=1/2 x0
(0,1 / 2) decreased, x > 1 / 2 increased
y=2x+8/x,y'=2-8/x^2=2[(x^2-4)]/x^2 ,y'=0
X = - 2, x = 2, (x = 0, y 'does not exist)
X