# One more problem for me

1. Oct 21, 2005

### Robokapp

Well, yesterday while doing homework i came across the following:

now...it looked simple. in fact it caused me severe collisions between the desk and my head...and here's why!

First chapter: Being lazy i figured why not use the chain rule. so...

dy/dx=(Ln x)*x^[(Ln x)-1]*1/x because derivative of Ln x=1/x

let's work it out. Ln x *x^(Ln x)*x^-1 * 1/x or simply by moving x on bottom

Ln x * x^Ln (x) *x^-2

okay...that is one answer. Now...the book has a formula for exponential derivatives that looks like dy/dx (a^b)= a^b *1/[b ln(a)] * b'

i don't think i'm right so i will not post what i did next becasue i'm not sure on the formula. anyway, skipping to chapter 3:

The book suggests logaritmic differentiation ( i think they call it that)

so: Chapter 3:

y=x^ Ln x
Ln y=Ln[x^(ln x)]
Ln y= Ln (x)^2 by log properties where ln x^2= 2 ln x
getting derivative:
1/y*dy/dx= 2Ln (x) *1/x and we know y=x^ ln x

dy/dx = 2Ln x * x^-1 * x^(ln x)

this is almoust what i got in first try...but not quite it.
The second one, which i got on papaer but not with me is also very close, i think it only has one less x on the bottom, but it's not the smae thing

can someone straighten this mess for me plz?

Natural logs are always doing this to me...

i'm looking for a reason why it doesn't work...and which one is correct. I'm sure that if i try again in a different way i'll get a different answer :D

Thank you
~Robokapp

2. Oct 21, 2005

### hypermorphism

The chain rule is used on a composition of functions, ie. to find the derivative of f(g(x)) with respect to x. What did you use for f and g such that f(g(x)) = x^(ln(x)) ? Note that if f(u) = u^n and g(x) = ln(x) as you seem to have used in your derivative, you only get f(g(x)) = ln(x)^n, not x^(ln(x)) as is required.
You should justify this step, as ln carries a limited domain. Show that y is always positive, and thus the domains of the two functions are the same.
Be careful! ln(x^2) = 2*ln(x), but you only have [ln(x)]^2. No further simplification is possible. The result of this differentiation is the correct one.

Last edited: Oct 21, 2005
3. Oct 21, 2005

### lurflurf

Use the power rule
$$\frac{d}{dx} \ u^v=v \ u^{v-1} \ \frac{du}{dx}+u^v \ \log(u) \ \frac{dv}{dx}$$
(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'

4. Oct 21, 2005

### Robokapp

Ln x *x^(Ln x)*x^-1 * 1/x

i did. well i said it's chain rule...i can't tell the difference between the two. i can do derivatives, but i can't tell what rule i'm using...i just...do it!

For some reason the power rule won't work if the power is not constant i think.

let's try.

x^2 => 2X right?
10^2 => 2*10^1*0, and that equals zero.

e^x=> e^x*x' so e^x=> e^x
but e^2 => 2e^1 * 0 => 0 which is not right. derivative of e^2 is not zero...but the exponential rule would make it be zero becasue of the * derivative of power.

i don't know...these rules contradict themselves depnding on scenario.

let's look at sin(xy)=0

cos(xy)*(y+x*dy/dx)=0 by chain rule

by product/power whatever rule:

consider (Sin (xy))^1

sin(xy)^-1*cos(xy)*(y+x*dy/dx)=0

so cot(xy)*(y+x*dy/dx)=0....and it's clearly wrong.

I'm getting dizzy...LOL

5. Oct 21, 2005

### lurflurf

All the rules are consistent.
x^2 => 2X right?
yes
10^2 => 2*10^1*0, and that equals zero.
yes
e^x=> e^x*x' so e^x=> e^x
yes
but e^2 => 2e^1 * 0 => 0 which is not right. derivative of e^2 is not zero...but the exponential rule would make it be zero becasue of the * derivative of power.
(e^2)'=0 as it is a constant
i don't know...these rules contradict themselves depnding on scenario.
let's look at sin(xy)=0
cos(xy)*(y+x*dy/dx)=0 by chain rule
by product/power whatever rule:
consider (Sin (xy))^1
sin(xy)^-1*cos(xy)*(y+x*dy/dx)=0
so cot(xy)*(y+x*dy/dx)=0....and it's clearly wrong.
should be
sin(xy)^(1-1)*cos(xy)*(y+x*dy/dx)=0
cos(xy)*(y+x*dy/dx)=0
since sin(xy)^0=1

I gave the power rule in a form that allows the base and exponent to both be nonconstant if needed.
Many calculus books are confusing in that they give many rules for the different cases

(u^a)'=a*u^(a-1)*u' (a constant)
and
(a^v)'=a^v*log(a)*v' (a constant)
are both special cases of
(u^v)'=v*u^(v-1)*u'+u^v*log(u)*v'

which is easily derived from
u^v=exp(v*log(u))
given derivatives of exp and log and the product rule
(u^v)=(exp(v*log(u)))'
derivative exp and chain rule
=exp(v*log(u))(v*log(u))'
product rule and u^v=exp(v*log(u))
=u^v(v(log(u))'+log(u)v')
derivative log
=u^v(vu'/u+log(u)v')
rearanging to final form
u^v=v*u^(v-1)*u'+u^v*log(u)*v'