Question

Write the first derivative of the given function. f(x)=[ ln(x) ]^2 / x^2

Answer 1

Use this:

d(f(x)/(g(x)))/ dx = (g(x)f'(x)-f(x)g'(x))/g(x)^2, where f'(x) = df/ dx denotes the derivative of f with respect to x.

Answer 2

I did that much I got
\[x ^{2}\times2lnx-[\ln(x)]^{2}\times2x \div x(x ^{4)}\]

Answer 3

i just don't see how to simply to the answer

Answer 4

yes - i slipped up - ill check it out on paper

Answer 5

okay thanks

Answer 6

use quotient rule

x^2 * 2lnx *(1/x) - 2x (lnx)^2
-------------------------
x^4

= 2x lnx - 2x(ln x )^2
--------------------
x^4

= 2xlnx(1 - lnx)
-----------
x^4

Answer 7

the answer is \[2(lnx)(1+lnx) \div x ^{3}\]

Answer 8

and where did 1/x go?

Answer 9

oh yes - i didnt cancel out the x and x^4 - that gives the answer you just quoted
the 1/x cancelled out with x^2 in the first line

Answer 10

ok and how did you get 1-lnx?

Answer 11

1-lnx is correct
thats the quotinet rule 1 + lnx is incorrect

Answer 12

quotient rule
if y = u/v
then dy/dx =[ vdu/dx - udv/dx ] / v^2