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differentiate y=x-lnx

i have to show this has 1 turning point. this involves differentiating and making it = 0.
can anyone tell me the method i should use for this please, its appriciated :biggrin:

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Reply 1

PhyMath

Could you show us some working so we know how much you can do?

Reply 2

Swayum

What exactly are you stuck on? You need to first differentiate the x and then differentiate the -lnx. Do you know their derivatives?

Reply 3

Totally Tom

gspot92

i have to show this has 1 turning point. this involves differentiating and making it = 0.
can anyone tell me the method i should use for this please, its appriciated :biggrin:

split it up into differentiating x and lnx

x diffs to 1 lnx diffs to 1/x

...

Reply 4

rbnphlp

gspot92

i have to show this has 1 turning point. this involves differentiating and making it = 0.
can anyone tell me the method i should use for this please, its appriciated :biggrin:

differentiating lnx gives 1/x

Spoiler

Reply 5

gspot92

i know that x goes to 1 and lnx to 1/x but does the - before the ln affect it in any way. do i then use the product rule to find dy/dx

Reply 6

rbnphlp

gspot92

i know that x goes to 1 and lnx to 1/x but does the - before the ln affect it in any way. do i then use the product rule to find dy/dx

what do you mean find dy/dx, 1-1/x is dy/dx

Reply 7

Totally Tom

\displaystyle \frac{d}{dx}af(x)=af'(x)

where a is a constant.

Reply 8

gspot92

if y=x-lnx
then u=x
du/dx=1
v=-lnx
what does dv/dx equal???

Reply 9

D-Day

There is no substitution involved. y=x differentiates to y'=1. y=-lnx differentiates to y'=-1/x. So y=x-lnx differentiates to y'=1-1/x. That's all she wrote.

Reply 10

Totally Tom

gspot92

if y=x-lnx
then u=x
du/dx=1
v=-lnx
what does dv/dx equal???

what
that
hell
are
you
on
with?

\displaystyle \frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)

Reply 11

AnonyMatt

gspot92

if y=x-lnx
then u=x
du/dx=1
v=-lnx
what does dv/dx equal???

Totally Tom

\displaystyle \frac{d}{dx}af(x)=af'(x)

where a is a constant.

a = -1.
Do you understand now?

Reply 12

DeeDub

This may help...

http://www.wolframalpha.com/input/?i=y%3Dx-lnx

Reply 13

maxfire

y = x - ln(x)

dy/dx = 1 - 1/x

dy/dx = 0 = 1 - 1/x

1 = 1/x

x = 1

y = x - lnx

y = 1 - ln1

y = 1

So the turning point is (1,1)

I THINK. I kind of rushed it.

Reply 14

Totally Tom

maxfire

y = x - ln(x)

dy/dx = 1 - 1/x

dy/dx = 0 = 1 - 1/x

1 = 1/x

x = 1

y = x - lnx

y = 1 - ln1

y = 1

So the turning point is (1,1)

I THINK. I kind of rushed it.

if they're retarded enough not to understand how to diff x-lnx then telling them how to do the next part is not the brightest idea just yet.

Reply 15

darkrosebud1991

maxfire

y = x - ln(x)

dy/dx = 1 - 1/x

dy/dx = 0 = 1 - 1/x

1 = 1/x

x = 1

y = x - lnx

y = 1 - ln1

y = 1

So the turning point is (1,1)

I THINK. I kind of rushed it.

what he said.

Reply 16

D-Day

Totally Tom

if they're retarded enough not to understand how to diff x-lnx then telling them how to do the next part is not the brightest idea just yet.

How nice of you to say :rolleyes:

Reply 17

gspot92

i don't get what any of you are on about, but ive kinda figured it out myself.

Reply 18

Totally Tom

gspot92

i don't get what any of you are on about, but ive kinda figured it out myself.

Good news everyone!

Reply 19

gspot92

maxfire

y = x - ln(x)

dy/dx = 1 - 1/x

dy/dx = 0 = 1 - 1/x

1 = 1/x

x = 1

y = x - lnx

y = 1 - ln1

y = 1

So the turning point is (1,1)

I THINK. I kind of rushed it.