# Find the derivative of the function y = lnx/1+lnx?

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You can solve this by writing

y = lnx/(1+lnx) as y = lnx(1+lnx)⁻¹

then use product & logaritmic formula

**d(uv) = udv + vdu**

** dln|u| = du/u**

dy = lnx(-1)(1+lnx)⁻²(dx/x) + (1+lnx)⁻¹dx/x

multiply x(1+lnx)²/dx both sides and simplify

x(1+lnx)²dy/dx = -lnx + (1+ lnx)

**dy/dx = 1/[x(1+lnx)²] **ans.

Since the function is a quotient, we'll use the quotient rule to differentiate the function with respect to x:

`dy/dx = ((ln x)'*(1 + lnx) - (ln x)(1 + ln x)')/(1+ln x)^2`

`dy/dx = (1/x + (lnx)/x - (ln x)/x)/(1+ln x)^2`

We'll reduce like terms from numerator:

`dy/dx = (1/x)/(1+ln x)^2`

**Therefore, the requested derivative of the function is **

`dy/dx = 1/(x*(1+ln x)^2)`