Question

Find any stationary values of thhe following curves and determine whether they are maxima or minima. Sketch the curves:

(a) y=x-Inx
(b) y=1/2x^2-In2x

Answer 1

Find the first derivative. Then set it to 0 to find the stationary point. To determine if it's a maxima or minima find the second derivative of the function.

Answer 2

lol what she said^

except when you say maxima and minima, do you mean the maximum / minimum
of f(x) or the maximum / minimum of f ' (x)

Answer 3

Ok, but how do you find the derivative of (a) for instance. I can find it when it's just Inx, but not when there's x-Inx.

Answer 4

\[\large y' = 1-\frac{1}{x} =>\frac{x-1}{x} \]

Answer 5

since derivative of x is 1.

Answer 6

How did you get to that answer?

Answer 7

And how would you get the second derivative?

Answer 8

Get to what answer? the derivative of x is 1 and the derivative of lnx is 1/x.

Then set it to zero to find the stationary point.

Answer 9

Well x - lnx, if you want to differentiate, you just do it like any function:

d/dx (x - lnx) = d/dx (x) - d/dx(lnx)

Answer 10

Um, Ok... I understand that now, @Mimi, you said that To determine if it's a maxima or minima find the second derivative of the function. So, how do you get the second derivative?

Answer 11

Just derive it again:

d/dx( 1 - 1/x) = 0 + 1/x^2 = 1/x^2

But i'm pretty sure you don't need the second derivative. I think you can just set the first derivative to 0:

0 = 1 - 1/x

x = 1

so plug this back into your original function:

y = 1 - ln(1)

y = 1 - 0

y = 1

so you're point is (1,1) and it must be a minimum since the range of the function is 1 to infinity.

Answer 12

I think you need to find the second derivative to determine the nature, like if it's a minimum or maximum.

Answer 13

Ok. Thanks! I understand now. I'm not ompletely sure of deriving still... What's the formula for deriving?

Answer 14

how would you derive (x + x^2)?

Answer 15

1 + 2x?
Not sure

Answer 16

Use the formula:
\[\large \frac{d}{dx} (ax+b)^{n} = n(ax+b)^{n-1} *a\]

Answer 17

you're right!

For derivatives, if you're adding/subtracting them, then you can just derive them individually

so for (x - lnx), you find the derivative of (x) which is 1, and you minus the derivative of lnx, which is 1/x

y = (x - lnx)

d/dx (y) = d/dx (x - lnx)

d/dx (y) = d/dx (x) - d/dx (lnx)

d/dx (y) = 1 - 1/x

Answer 18

But, how do you work with derivatives of In?
That's what I'm stuck on

Answer 19

The derivative of lnx => 1/x

Answer 20

You just need to remember it.

Answer 21

derivative of lnx = 1/x

I mean, if you want to see the proof for it, I could probably type it up... but it's really long

Answer 22

No, that's fine... The Inx is fine, it's just, what if it was 2In4x and you had to derive that?

Answer 23

well then that just becomes

2(1/4x)(4)

= 8(1/4x) = 8/4x = 2/x

you have to chain rule the 4x, which will pop out a 4.

Answer 24

Ok.. How about In 1/(3x+1)?

Answer 25

It was a question I worked with recently, and couldn't solve..

Answer 26

Just change it to:
\[(3x+1)^{-1} \]
then differentiate it.

Answer 27

No, it's
In (3x+1)^-1

Answer 28

woops, sorry

Answer 29

same idea, just imagine the thing inside natural log as one whole chunk:

y = ln(3x+1)^-1

y'= [1/(3x+1^)-1] (-(3x + 1)^-2) ( 3)

y'= (3x+1)(-1/(3x + 1)^2) (3)

do you see how you can reduce the 3x + 1 with the 1/(3x + 1)^2

y' = -1/(3x+1) (3)

y' = -3/(3x + 1)

Answer 30

Oh Ok. Thanks! :)