wat more can i do with this derivative?

Question

lnx/(x^2+1) is f and for f' i got {[(x^2-1)/x] -lnx*2x}/(x^2+1)^2 did i do this rite?

Answer 1

Given:
lnx / (x²+1)

Use the formula: dy/dx = [ V du/dx - U dv/dx ] / V²
Let U = lnx then du/dx = 1/x
and V = (x²+1) then dv/dx = 2x
and V² = (x²+1)²

=> dy/dx = [ (x²+1)(1/x) - (lnx)(2x) ] / (x²+1)²
=> dy/dx = [ (x+1/x) - (lnx)(2x) ] / (x²+1)²

Answer 2

Almost:

{[(x^2+1)/x] -lnx*2x}/(x^2+1)^2

Answer 3

(lnx)/(x^2 + 1)

(((x^2 + 1)(lnx)') - ((x^2 + 1)'(lnx)))/((x^2 + 1)^2)
(((x^2 + 1)(1/x)) - 2xlnx)/((x^2 + 1)^2)
(((x^2 + 1)/x) - (2xlnx))/((x^2 + 1)^2)

Only one equation, where did you get (x^2 - 1)/x

I continued, and here is what i got.

(((x^2 + 1)/x) - ((2x^2(lnx))/x)((x^2 + 1)^2)
((x^2 + 1 - 2x^2)/x)/((x^2 + 1)^2)
((-x^2 + 1)/x)/((x^2 + 1)^2)
((-x^2 + 1)/x)/(((x^2 + 1)^2)/1)
((-x^2 + 1)/x)*(1/((x^2 + 1)^2)))
(-x^2 + 1)/(x(x^2 + 1)^2)

-(x^2 - 1)/(x(x^2 + 1)^2)

Answer 4

f = [ln x] / [x^2 + 1]

Quotient rule

f' = [(1/x) * (x^2 + 1) - (ln x) * (2x)] / [x^2 + 1]^2

... = (1/x) / (x^2 + 1) - (2x ln x) / (x^2 + 1)^2

... = 1/(x^3 + x) - (2x ln x) / (x^4 + 2x^2 + 1)

It depends what form you want to end up with :)

Answer 5

Hmm, (bottom*dtop - top*dbottom)/bottom^2

(((x^2+1)/x)-2x*ln(x)
---------------------------------------
(x^2 + 1)^2

Let's clear fractions a little by multiplying the top and the bottom by x:

(x^2 + 1) -2x^2*ln(x)
------------------------------
(x^2 + 1)^2

Gathering like terms on the top we can get:

(2ln(x) +1)x^2 + 1
-------------------------------
(x^2 + 1)^2

Nope, I don't see any benefit to this attempt at simplification. I would stop at the one before this.

Answer 6

f(x) = ln (x) / (x^2 + 1) ----> quotient rule

f'(x) = [ (x^2 + 1)*(1 / x) - (ln(x))*(2x) ] / (x^2 + 1)^2

= [ x + (1/x) - 2x*ln (x) ] / (x^4 + 2x^2 + 1)