How would I find dy/dx of y = logx / (1 + logx) ?

Answer 1

Quotient rule, and then chain rule.

The first thing to deal with on the right side is the fraction. Use the quotient rule there: "low d-high minus high d-low over the square of what's below," as my calc teacher taught us.

(1+log x) d/dx (log x) - (log x) d/dx (1+log x)
---------------------------------------------------------
(1+log x)^2

Now use the chain rule and finish out those remaining derivatives. d/dx (log x) is 1/(x ln 10). (Unless you meant ln x, in which case it's simply 1/x.) d/dx (1+log x) is the same thing, because 1 is a constant.

(1+log x)/(x ln 10) - (log x)/(x ln 10)
---------------------------------------------
(1+log x)^2

This doesn't look all too hard to simplify (the log x's in the top will cancel, for instance), so I'll leave that to you. The calculus is done.

Answer 2

well you have to know about the chain rule for differentiation. that says that when h(t) = f(t)*g(t)
then h' = f'g + fg'

If you dont know that you can not solve the dy/dx

The next thing you need to know is that d log(x) /dx = 1/x
and also that d x^-1/dx = -1/x^2

that is all

set y = h. f = log x ; g = 1/(1+ log x)
and do what i just explained.

Answer 3

[(1+logx) {d/dx(logx)} - logx {d/dx(1+logx)}] / (1+logx)(1+logx)

Answer 4

use the quotient rule,

remember that log x = ln x / ln 10

Answer 5

ask a teacher when school starts