How do I compute the derivative of ln (7x+1)^1/2 (3x^2+x)^5 / (x^2-3)^3 e^2x?

[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

 

[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...

 

[nrqed]

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

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.

 

[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?

 

[neutrino]

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

 

[HallsofIvy]

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

 

[nanai]

Thanks everyone