Using derivative to find the extreme value of y = 2x / ln x

Using derivative to find the extreme value of y = 2x / ln x

y'=[(2x)'*lnx-2x*(lnx)']/(lnx)^2
=(2*lnx-2x*1/x)/(lnx)^2
=2(lnx-1)/(lnx)^2
Let y '= 0
Then lnx-1 = 0
x=e
Domain x > 0, LNX is not equal to 0
So x > 0 and not equal to 1
x>e,lnx-1>0,{lnx}^2>0
So y '> 0, y increases monotonically
one

Finding the second derivative 6xlnx + 5x of y = x ^ 3 ln x

y'=3x^2 ln x+x^3*1/x=3x^2 ln x+x^2
y''=6xlnx+3x+2x=6xlnx+5x

Is the derivative of Ln (5x) 5 / x?

No, ln (5x) '= 5 / (5x) = 1 / X. we can also see that ln (5x) = LN5 + LNX, so the derivative is 1 / X

How to find the derivative based on LN

Inx=1/x
aInx=a/x