How Do You Differentiate Products of Logarithms?

kuahji
Messages
390
Reaction score
2
[SOLVED] derivative with logarithms

Ok, so the problem problem probably isn't as bad as I'm making it, either that or its because its getting late & my brain just isn't functioning.

Find the derivative of y with respect to r.
y=[tex]log _2 \left( r \right)[/tex] * [tex]log _4 \left( r \right)[/tex]

The first thing I thought to do was use the product rule which yielded
y'=[tex]log _2 \left( r \right)[/tex]*(1/ln2)(1/r)+[tex]log _4 \left( r \right)[/tex](1/ln4)(1/r)

I then changed the [tex]log _2 \left( r \right)[/tex] to (ln r/ln2) & did the same to the other logarithm which created a complex fraction that I condensed down to
(ln r)/(r ln 2 * ln 2) + (ln r)/(r ln 4 * ln 4)

Cross multiplying gave
[(ln r)(ln 4)^2+ (ln r)(ln 2)^2]/[r (ln 2)^2 (kn 4)^2]

Here is where I'm stuck, because the book likes (2 ln r)/(r ln 2 * ln 4).

The book also showed different steps. To begin with it shows making the logs into ln.
which gives
y=(ln r)^2/(ln 2 * ln 4)
then take the derivative, but here is where I'm a bit lost. It shows
y'=1/(ln2 *ln 4) * (2 ln r) * (1/r)
I can see where the (2 ln r) & the (1/r) come from the chain rule, but not so much the 1/(ln2 *ln 4) part.
 
on Phys.org
y= (ln r/ln 2)*(ln r/ln 4) = (ln r)^2/(ln 2*ln 4), as in the book. The denominator D is a const, and just keep it like that. So,

y' = 2(ln r)(1/r)/D.
 
That'd probably eventually simplify to the correct answer, but I think your teacher wanted you to apply the change-of-base formula at the start and then use the chain rule as follows:

[tex]\frac{dy}{dr} \left ( \frac{\ln^2 r}{\ln 2 \ln 4} \right ) = \frac{2\ln r}{r \ln 2 \ln 4}[/tex]
 
foxjwill said:
That'd probably eventually simplify to the correct answer, but I think your teacher wanted you to apply the change-of-base formula at the start and then use the chain rule as follows:

[tex]\frac{dy}{dr} \left ( \frac{\ln^2 r}{\ln 2 \ln 4} \right ) = \frac{2\ln r}{r \ln 2 \ln 4}[/tex]

How is this any different from what I'd given?

Also, [tex]( \ln^2 r)[/tex] is not a standard notation.
 
Shooting star said:
The denominator D is a const, and just keep it like that.

Right, duh... :) thanks.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K