Differentiation help needed

Original post by SteveMcs

it is wrong: you cannot just use the chain rule. Have you done logarithmic differentiation?

Example: can you differentiate

x^x

(edited 9 years ago)

Reply 3

Original post by tombayes

it is wrong, very wrong in fact: you cannot just use the chain rule. Have you done logarithmic differentiation?

Example: can you differentiate

x^x

No, i've never done that :s-smilie:

Reply 4

Original post by SteveMcs

No, i've never done that :s-smilie:

How would you differentiate:

x^x

Reply 5

Oh hang on... is this like x^x= e^xlogx ??

Reply 6

happysmile

we started differentiation today :/

Reply 7

Original post by SteveMcs

Oh hang on... is this like x^x= e^xlogx ??

yes

Reply 8

Original post by tombayes

yes

So i'm guessing i just do my equation the same with my value of x, thank you!

Reply 9

Original post by SteveMcs

So i'm guessing i just do my equation the same with my value of x, thank you!

How would you write

(x^2+3)^x

in the same way as you did

x^x=e^{x\ln x}

?

Also, going back to your first post, the derivative of

x^n

nx^{n-1}

but this only works for constant values of

n

so you can't differentiate

x^x

in the same way.

This is the first of a few reasons why your method was incorrect.

Reply 10

So mine would be e^{xlog(x^2 +3)}so (x²+3)^x * d/dx(log(x²+3))

(edited 9 years ago)

Reply 11

Original post by SteveMcs

So i'm guessing i just do my equation the same with my value of x, thank you!

it would be quite difficult to it that way.

This is how I would differentiate

x^x

Let

y=x^x

then

log(y)=xlog(x)

where

log(x)

is the natural log (obviously)
now differentiating we get:

\frac{1}{y}\frac{dy}{dx} =1+log(x)

by chain rule
now rearranging and remembering

y=x^x

we get

\frac{dy}{dx}=x^x(1+log(x))

Now can you apply this method to your question?

Reply 12

L'Evil Fish

Let y = (x^2+3)^x or whatever

Take logs, bring x down

Diff wrt x

Multiply through by y, you know what y is
Voila

Or so I assume...

Reply 13

Original post by SteveMcs

So mine would be e^{xlog(x^2 +3)}so (x²+3)^x * d/dx(log(x²+3))

That's not right. Can you post your working?

Your edit is closer, but still not right.

Reply 14

Original post by notnek

That's not right. Can you post your working?

Your edit is closer, but still not right.

Sorry, ywah know now it's wrong... i didn't fully understand until l'evel fish's post. makes perfect sense now :smile:

thanks everyone!

Reply 15

Original post by tombayes

it would be quite difficult to it that way.

Not really. It can be done using either methods and both have a similar number of steps.

Reply 16

Original post by SteveMcs

Sorry, ywah know now it's wrong... i didn't fully understand until l'evel fish's post. makes perfect sense now :smile:

thanks everyone!

OK. If you're still interested in the method you were using :

\displaystyle (x^2+3)^x = e^{x\ln{(x^2+3)}}

Using the product rule:

\displaystyle \frac{d}{dx}\left(x\ln{(x^2+3)} \right) = \ln{(x^2+3)} + x\frac{2x}{x^2+3}

and go from there.

(edited 9 years ago)

Reply 17

Just to check my new answer...

let y=(x²+3)^x
log both sides
logy=xlog(x²+3)
differentiate implicitly
1/y dy/dx = 2x²/(x²+3) + log(x²+3)
since y=(x²+3)^xdy/dx = (x²+3)² * (2x²/(x²+3) + log(x²+3))

Reply 18