# logorithmic differentiation

• Dec 9th 2009, 09:24 AM
Evan.Kimia
logorithmic differentiation
http://www.webassign.net/cgi-bin/sym...%2A%287%20x%29

Heres what i did:

used the laws of logs to get this:

7x ln (sqrt(x))

ln(y) = 7x ln (sqrt(x))

start solving by implicit differentiation:

y'/y = 7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7

final = y'=y(7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7)

I think i know what i did wrong, i used ln(sqrt(x)) as 1 function instead of 2. This is a Web Assign and im about out of guesses. If I would apply the product rule between 7x and ln then apply it in another product rule with sqrt(x) would i get the right answer?
• Dec 9th 2009, 09:30 AM
TheEmptySet
Quote:

Originally Posted by Evan.Kimia
http://www.webassign.net/cgi-bin/sym...%2A%287%20x%29

Heres what i did:

used the laws of logs to get this:

7x ln (sqrt(x))

ln(y) = 7x ln (sqrt(x))

start solving by implicit differentiation:

y'/y = 7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7

final = y'=y(7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7)

I think i know what i did wrong, i used ln(sqrt(x)) as 1 function instead of 2. This is a Web Assign and im about out of guesses. If I would apply the product rule between 7x and ln then apply it in another product rule with sqrt(x) would i get the right answer?

the equation can be further simpilifed before you take the derivative:

$\ln(y)=7x\ln(\sqrt{x})=\frac{7x}{2}\ln(x)$

You could also note this from the beginning

$y=(\sqrt{x})^{7x}=(x^{1/2})^{7x}=x^{\frac{7x}{2}}$

Then take the log

So then we get

$\frac{y'}{y}=\frac{7}{2}\ln(x)+\frac{7}{2}=$ ...

I hope this helps