What is the solution for, d/dx(logx/x)

d/dx(logx/x)

= [d/dx(logx)*x-logx*d/dx(x)]/x^2

= [1/x*x-logx]/x^2

= [1-logx]/x^2

= 1/x^2-logx/x^2 Ans.

Just use the product rule,

y' = u'v + v'u (the quotient rule is good as well)

let u = log(x ) u' = 1/x
v = 1/x v' = -1/x^2

y' = 1/x * 1/x + -1/x^2 * log(x)
= 1/(x^2) * (1 - log(x))

I am assuming you mean d/dx(log(x)/x)...if so then use quotient rule..
(u/v)' = (vdu - udv)/v^2

d/dx(log(x)/x) = [x(1/(xln(10))) - log(x)] / x^2
= (1 - ln(10)log(x)) / (ln(10)x^2)

d/dx (logx/x) = (d/dx(logx) * x - logx * d/dx(x)] / x^2
. . . . . . . . . . = (1/x * x - logx) / x^2
. . . . . . . . . . = (1 - logx) / x^2

Mαthmφm

Use the quotient rule

(x/x – logx)/x² = (1 – logx)/x² Assuming that log x is the natural log

(1 – lnx)/(x²ln(10)) if it is the common log

d/dx(logx/x)
= [x*1/x - log(x)]/x^2
=(1-log(x)/x^2

d/dx(logx/x)

=> 1/x * 1/x - logx * /x^2

=> 1/x^2 - logx * /x^2

=> (1- logx) * /x^2

FOR WHAT????