1. ## First and Second Derivative with ln, Help please

Hey, I am having trouble finding either the first or second derivative of a problem. Could someone show me how to solve both and simplify? It will be really helpful with my other homework problems.

f(x)= x(ln x)^2

I need to find f '(x) and f ''(x). Thank you

2. Originally Posted by lostsheep
Hey, I am having trouble finding either the first or second derivative of a problem. Could someone show me how to solve both and simplify? It will be really helpful with my other homework problems.

f(x)= x(ln x)^2

I need to find f '(x) and f ''(x). Thank you
First the product rule, then the chain rule.

$f(x) = x \cdot ln^2(x)$

$f^{\prime}(x) = (x)^{\prime} \cdot (ln^2(x)) + x \cdot (ln^2(x))^{\prime}$

Now,
$(ln^2(x))^{\prime} = 2 \cdot ln(x) \cdot \frac{1}{x}$, so
$f^{\prime}(x) = 1 \cdot (ln^2(x)) + x \cdot 2 \cdot \frac{ln(x)}{x}$

$f^{\prime}(x) = ln^2(x) + 2ln(x)$

Then
$f^{\prime \prime}(x) = 2 \cdot ln(x) \cdot \frac{1}{x} + 2 \cdot \frac{1}{x}$

$f^{\prime \prime}(x) = \frac{2ln(x)}{x} + \frac{2}{x}$

-Dan

3. Originally Posted by topsquark
First the product rule, then the chain rule.

$f(x) = x \cdot ln^2(x)$

$f^{\prime}(x) = (x)^{\prime} \cdot (ln^2(x)) + x \cdot (ln^2(x))^{\prime}$

Now,
$(ln^2(x))^{\prime} = 2 \cdot ln(x) \cdot \frac{1}{x}$, so
$f^{\prime}(x) = 1 \cdot (ln^2(x)) + x \cdot 2 \cdot \frac{ln(x)}{x}$

$f^{\prime}(x) = ln^2(x) + 2ln(x)$

Then
$f^{\prime \prime}(x) = 2 \cdot ln(x) \cdot \frac{1}{x} + 2 \cdot \frac{1}{x}$

$f^{\prime \prime}(x) = \frac{2ln(x)}{x} + \frac{2}{x}$

-Dan
in the original problem it is stated f(x) = x(ln x)^2

so (ln x) part is squared, is this the same as ln^2 (x)?

I am not sure, but thanks for the help

4. Originally Posted by lostsheep
in the original problem it is stated f(x) = x(ln x)^2

so (ln x) part is squared, is this the same as ln^2 (x)?

I am not sure, but thanks for the help
Yes they are the same thing with different notations,
just as sin^2(x)=(sinx)^2.