Find derivative of y. y= ln (1 + √x) / (x^3)

[HellRyu]

Homework Statement

Hi guys, I've got :

$$y= ln ( (1 + √x) / x^{3})$$


2. The attempt at a solution
I honestly don't know where to go from here, I tried getting the ln of each of them.


$$y = ln 1 +ln√x - ln x^{3}$$


Am I doing it write? If not, how am I suppose to work this problem out? If so, where do I go from here?

 

[MarneMath]

First note, log(1+sqrt(x)) does not equal log(1) + log (sqrt(x)). You're thinking about this property:

log(ab) = log(a) + log(b). So, what you should have is this mess below:

$y = \ln{(1+\sqrt{x})} - \ln{x^3}$

On the first part, use the chain rule, on the second part use the chain rule. Show your work and we'll see where you go astray.

 

[HellRyu]

First note, log(1+sqrt(x)) does not equal log(1) + log (sqrt(x)). You're thinking about this property:

log(ab) = log(a) + log(b). So, what you should have is this mess below:

$y = \ln{(1+\sqrt{x})} - \ln{x^3}$

On the first part, use the chain rule, on the second part use the chain rule. Show your work and we'll see where you go astray.

O.K. so I started doing the chain rule for the first one and got:

$$(1/(1 + √x) )(1/(2√x) )$$

Is it right so far?

EDIT: I did the second one and got:

$$(1/(x^{3}) ) (3x^{2} )$$

 

[HellRyu]

ok I got used the chain rule and got

$$[(1/2x^{-1/2})/(1 + √x)] - [(3x^2)/(x^3)]$$

then

$$1/[ (2√x) + 2x ] - 3/x$$

How do I go from here to get the answer :

$$(-6 -5√x)/[2x(1 + √x) ]$$ ?

 

[SammyS]

ok I got used the chain rule and got

$$[(1/2x^{-1/2})/(1 + √x)] - [(3x^2)/(x^3)]$$

then

$$1/[ (2√x) + 2x ] - 3/x$$

How do I go from here to get the answer :

$$(-6 -5√x)/[2x(1 + √x) ]$$ ?

Find a common denominator & combine fractions.