# Ln derivative

1. Apr 22, 2006

### nanai

The question is to compute the derivative of
ln (7x+1)^1/2 (3X^2+x)^5
(x^2-3)^3 e^2x
the ln is for both the numerator and denominator

I tried the chain rule and came up with this

1
(7x+1)^1/2 (3x^2+x)^5
(x^2-3)^3 e^2x

and finally 6x(x^2-3)^2 2e^2x
7/2(7x+1)^-1/2 30x(3x^2+x)^4

2. Apr 22, 2006

### moose

You may want to try to apply some logarithm properties to that to simplify it.

For example, lna/b is the same thing as lna - lnb. lna^b = blna. lna*b = lna + lnb.....

3. Apr 22, 2006

### nrqed

Are you sayingthat your final expression is your final result for the derivative? That is not correct.

As Moose said, use first properties of the ln to write this as a *SUM* of 4 terms and bring down the exponents using ln(a^n)= n ln(a). Also, use $ln ( e^{2x}) = 2x$. Only after doing all that you should take the derivative and it will be quite simple.

4. Apr 23, 2006

### nanai

ok I used the ln property and here is what I got.

ln(7x+1)^1/2+ln(3x^2+x)^5-ln(x^2-3)^3+lne^2x

1/2ln(7x+1) +5ln(3x^2+x)-3ln(x^2-3)+2x

and finally 1/2*7/7x+1+5*6x/3x^2+x -3*2x/x^2-3 + 2

am I right?

5. Apr 23, 2006

### neutrino

Apart from the '+1' missing from numerator of the third term(in the final answer), everything else seems to be right.

6. Apr 23, 2006

### HallsofIvy

It's -2x, not +2x (e2x was in the denominator of the fraction.)
And, of course, you'll want to simplify those fractions.

7. Apr 23, 2006

### nanai

Thanks everyone