The derivative of Y is Xe ^ x (x times e to the power of x). Find y You can also give the results directly

The derivative of Y is Xe ^ x (x times e to the power of x). Find y You can also give the results directly

(x-1)e^x+C

1. Y = [(B / a) to the power of x] times [(x / b) to the power of a] times [(A / x) to the power of B] 2. Y = (x-1) (X-2) ^ 2... (x-n)^ The extended problem y = x (x-1) (X-2)... (x-n) finding f '(n) aln|x/b|+bln|a/x| Is the derivative equal to ln|a / x| + ln|b / x|? Shouldn't it be equal to a / X-B / x? Why does ln exist?

Logarithm of absolute value
1.ln|y|=xln|b/a|+aln|x/b|+bln|a/x|
y'/y=ln|b/a|+ln|a/x|+ln|b/x|=ln|b|-ln|a|+ln|a|-ln|x|+ln|b|-ln|x|=2(ln|b/x|)
y'=2(ln|b/x|)y
2.ln|y|=ln|x|+ln|x-1|+ln|x-2|+...+ln|x-n|
y'/y=1/x+1/(x-1)+1/(x-2)+...+1/(x-n)
y'=[1/x+1/(x-1)+1/(x-2)+...+1/(x-n)]y=[1/x+1/(x-1)+1/(x-2)+...+1/(x-n+1)]y+x(x-1)(x-2)...(x-n+1)
Substitute x = n to get
y'=n(n-1)(n-2)...1=n!

How to find the derivative of the curve y = - x times e to the x power?

The derivative is equal to the x power of - (1 + x) e

Take the derivative of ln a multiplied by the x power of A Please write down the process

(lna*a^x)!= (lna)!* a^x+lna*(a^)!= 0+(lna)^2*a^x=(lna)^2*a^x

Z = ln (Y / x) how to find the partial derivative Find the partial derivative of Z = x ^ (XY)

z'x=(-y/x^2)/(y/x)=-1/x
z'y=(1/x)/(y/x)=1/y
dz=z'xdx +z'ydy
u=ln(x^2+y^2+z^2)
u'x=2x/(x^2+y^2+z^2)
u'y=2y/(x^2+y^2+z^2)
u'z=2z/(x^2+y^2+z^2)
du=2x/(x^2+y^2+z^2) dx + 2y/(x^2+y^2+z^2) dy + 2z/(x^2+y^2+z^2) dz

Given z = ln (x + y ^ 2), find the second partial derivative of Z

∂z/∂x=1/(x+y^2),∂z/∂y=2y/（x+y^2）
∂^2z/∂x^2=-1/（x+y^2）^2,∂^2z/∂y^2=[2（x+y^2）-4y^2]/（x+y^2）^2
∂^2z/∂x∂y=-2y/(x+y^2)^2,∂^2z/∂y∂x=-2y/(x+y^2)^2