How to derive y = ln (1 + x) ^ 1 / X

How to derive y = ln (1 + x) ^ 1 / X

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

Y = (5x ^ 2 + 1) ^ 10 derivation of composite function

y=[10(5x^2+1)^9]*(5x^2+1)'
=[10(5x^2+1)^9]*10x
=100x(5x^2+1)^9

Derivative of compound function x / z = ln (Z / y) partial derivative of Z to x what is the answer

x/z=ln(z/y),
Differential:
（zdx-xdz)/z^2=y/z*(ydz-zdy)/y^2=(ydz-zdy)/(yz),
∴yzdx-xydz=yzdz-z^2dy,
∴z'=yz/(xy+yz)=z/(x+z).

The function y = [cos (x - π / 12)] ^ 2 + [sin (x + π / 12)] ^ 2-1 has a period of___ Yes____ (odd / even) function

y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1
=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2-1
=[cos(2x-π/6)-cos(2x+π/6)]/2
=(1/2)*(-2)sin2xsin(-π/6)
=(1/2)sin2x
So fill in: π, odd

What is the minimum positive period of the function y = sin ^ 4-cos ^ 4?

y=sin^4x-cos^4x
=(sin ² x+cos ² x)(sin ² x-cos ² x)
=sin ² x-cos ² x
=-2cos2x
Minimum positive period T = 2 π / 2 = π

Senior one mathematics: what is the minimum positive period of the function y = cos ＾ 4-sin ＾ 4?

y=cosx＾4-sinx＾4
=1（cosx＾2-sinx＾2）
=cos2x
2pai/2=pai