What is the derivative of the natural log ... X ln X ^ 1/2?

Question

I understand that when you have a simple "ln x" that the answer is ..

1/x * dx

But with the element in front of it, I'm not sure ... I got 1/2 :-/

Answer 1

f(t) = t sqrt( ln (t))

let ln t = u

f(t) = t sqrt(u)

f'(t) = sqrt(u) +{ t / 2sqrt(u) }(du/dt)

The above using the product rule.

f'(t) = sqrt( ln t) + [ t / 2 sqrt (ln (t)) ] 1/t]

f'(t) = sqrt( ln t ) + 1/ 2t sqrt(ln t)

Note: The derivative of sqrt(something) = 1/2sqrt(something)

Answer 2

y = x ln x^1/2 = (1/2)x lnx (I brought the 1/2 down by the log power rule)

this is a product

if y = uv, then y' = u'v + uv'

so...

y' = (1/2)ln x + (1/2)x (1/x)

y' = (1/2)(ln x + 1)

Answer 3

You must use the product rule then the chain rule

product rule

x'x+xx'

chain rule

(lnx^1/2)'=

lnu=1/u and (x^1/2)' =1/2x^1/2+1=1/2x^3/2

and then pluge that answer into lnu you will get ln(1/2x^3/2) which is the derivative of lnx^1/2

so then plugging that into the chain rule u did earlier

=1*lnx^1/2+ln(1/2x^3/2)x is the answer unsimplified