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

  • Thread starter HellRyu
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  • #1
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Homework Statement



Hi guys, I've got :

[tex] y= ln ( (1 + √x) / x^{3})[/tex]


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.


[tex]y = ln 1 +ln√x - ln x^{3} [/tex]


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

Answers and Replies

  • #2
MarneMath
Education Advisor
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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:

[itex] y = \ln{(1+\sqrt{x})} - \ln{x^3} [/itex]

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.
 
  • #3
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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:

[itex] y = \ln{(1+\sqrt{x})} - \ln{x^3} [/itex]

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:

[tex] (1/(1 + √x) )(1/(2√x) ) [/tex]

Is it right so far?

EDIT: I did the second one and got:

[tex] (1/(x^{3}) ) (3x^{2} )[/tex]
 
Last edited:
  • #4
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ok I got used the chain rule and got

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

then

[tex] 1/[ (2√x) + 2x ] - 3/x [/tex]

How do I go from here to get the answer :

[tex](-6 -5√x)/[2x(1 + √x) ] [/tex] ?
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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ok I got used the chain rule and got

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

then

[tex] 1/[ (2√x) + 2x ] - 3/x [/tex]

How do I go from here to get the answer :

[tex](-6 -5√x)/[2x(1 + √x) ] [/tex] ?
Find a common denominator & combine fractions.
 

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