easy log question, help (1 Viewer)

[BoganBoy]

This is probably really easy but i dont know where to start.

Differentiate
y=x^x

Can someone please help? Thanks

 

[香港!]

This is probably really easy but i dont know where to start.

Differentiate
y=x^x

Can someone please help? Thanks

y=x^x
ln y=ln x^x=xlnx
(1\y)y'=lnx+1
y'=y(lnx+1)

 

[BoganBoy]

can you please explain how you got (1\y)y'? what does it represent? Thanks

 

[香港!]

Implicit Differentiation:
http://mathworld.wolfram.com/ImplicitDifferentiation.html

 

[Dumsum]

Implicit Differentiation rocks my socks.

As an addendum, I'm pretty sure this is beyond 2u scope.

 

[mojako]

Implicit Differentiation:
http://mathworld.wolfram.com/ImplicitDifferentiation.html

HK attack!

if u dont want to use the method they do in Hong Kong,
y=x^x
y=e^(x lnx)

if u=x lnx,
du/dx = 1 lnx + x 1/x = lnx + 1

y=e^(x lnx)
y = (lnx + 1) e^(x lnx)

hihihihihi

*flee!*

*comes back*
anyway u wont get this kind of question

*steal!*
*flee!*

 

[香港!]

@_@
how dare u attack me??
Ultima!!

 

[mojako]

@_@
how dare u attack me??
Ultima!!

doesnt work against me
hihihihihihihi

 

[klaw]

HK attack!

if u dont want to use the method they do in Hong Kong,
y=x^x
y=e^(x lnx)

if u=x lnx,
du/dx = 1 lnx + x 1/x = lnx + 1

y=e^(x lnx)
y = (lnx + 1) e^(x lnx)

hihihihihi

*flee!*

*comes back*
anyway u wont get this kind of question

*steal!*
*flee!*

isn't substitution calculus also beyond 2U?

 

[Jago]

very much so.

 

[mojako]

isn't substitution calculus also beyond 2U?

well.. depends on what substitution it is
i think 2u ppl shud be able to differentiate
y=e^(x lnx)
or maybe simpler ones like y=e^sinx

but the trick with this q is to get
y=x^x
y=e^(x lnx)

and that shudnt be in 2u

 

[香港!]

if u have
y=e^x
when u differentiate it would be
y'=e^x

in the process u let u=x already
so that's also "substitution" LoL

 

[mattchan]

well.. depends on what substitution it is
i think 2u ppl shud be able to differentiate
y=e^(x lnx)
or maybe simpler ones like y=e^sinx

but the trick with this q is to get
y=x^x
y=e^(x lnx)

and that shudnt be in 2u

They teach this in the 2unit Cambridge Book, in the last section of this topic i think.

 

[klaw]

They teach this in the 2unit Cambridge Book, in the last section of this topic i think.

teach x^x=e^(x lnx) or how to differentiate e's?

 

[mattchan]

How to change x^x into an exponential

 

[acmilan]

The derivative of a^x is a 2 unit topic, thus all 2 unit students should know that a^x = e^xlna.

 

[.ben]

i'm doing 3u and how come we haven't done that yet?

 

[Dumsum]

i'm doing 3u and how come we haven't done that yet?

HSC: 2006


You'll do it next year. Next term if you're lucky