# Derivative with logarithms

1. Jan 26, 2008

### kuahji

[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=$$log _2 \left( r \right)$$ * $$log _4 \left( r \right)$$

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

I then changed the $$log _2 \left( r \right)$$ 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.

2. Jan 27, 2008

### Shooting Star

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.

3. Jan 27, 2008

### foxjwill

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:

$$\frac{dy}{dr} \left ( \frac{\ln^2 r}{\ln 2 \ln 4} \right ) = \frac{2\ln r}{r \ln 2 \ln 4}$$

4. Jan 27, 2008

### Shooting Star

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

Also, $$( \ln^2 r)$$ is not a standard notation.

5. Jan 27, 2008

### kuahji

Right, duh... :) thanks.