Question

Derive y=(1/2)x^2 - In2x

Answer 1

First, whats the derivative of (1/2) x^2?

Answer 2

x?

Answer 3

yes! now whats the derivative of -ln2x

Answer 4

-1/x?

Answer 5

remember the chain rule ;)

Answer 6

derivative of -ln2x = (-1/2x)(2) <--- chain rule the 2x

Answer 7

now just put them together:

x - 1/2x (2)

Answer 8

Why is (-1/2x)(2) not the same as (-1/x)?

Answer 9

oh i'm sorry you're absolutely right. My mind slipped for a second :(

Answer 10

but you're getting the hang of it and even correcting me! so thats what matters right! :D

Answer 11

I'm having a wrong answer with the stationary value then :/ Can you see if you get it?

Answer 12

I thought my deriving was wrong... Must be the stationary value.

Answer 13

what was the original question?

Answer 14

The same as the last, the second question I asked there. Finding the stationary value + maxima and minima

Answer 15

ok we know the derivative is x - 1/x right?

so the stationary value is just when the derivative = 0, so we set x - 1/x = 0

0 = x - 1/x

x = 1

my Calc 1 is a little rusty, I'll have to get back to you on how to find maximums and minimums :(

Answer 16

Ok, so is the text book wrong? because I got 1 aswell.
1/2-In2
is what is says

Answer 17

ohhh, because you take x=1 and plug it back into the original equation:

y = (1/2) (x)^2 - ln(2x)

y = (1/2) (1) - ln2

y = 1/2 - ln2

so the stationary POINT is (1, 1/2 - ln2)

1/2 - ln2 is the height of the point. Just looking at it, I would intuitively say its probably the minimum.

Answer 18

Oh! Ok. That makes more sense. Thank you!!

Answer 19

Yea no prob. I'm gonna head out now, so if you have anymore calculus questions ask me 2m :D

Answer 20

Sure. Thanks again!