Calc2 - Dx with ln

littlejodo · #1 $Old$ January 22nd, 2009, 04:19 PM

Find dr/dx if

r = (lnx)/((x^2)lnx^2) + ln(1/x)^3

r' = lnx(x^2 lnx^2)^-1 + 3ln(1/x)

r' = 1/x(x^2 lnx^2)^-1 - lnx(x^2 lnx^2)(2x/x^2) + 3x --> first term I used chain and product rules, but I'm not sure if I carried them out correctly.

then I tried to combine stuff...

r' = 1/(x(x^2lnx^2) - lnx(x^2 lnx^2)(2x/x^2) + 3 --> is this the correct answer? None of the examples I saw in class still had "ln" in the final answer so I wasn't sure if I messed something up... (a high probability scenario).

Thanks!

Danny · #2 $Old$ January 22nd, 2009, 04:28 PM

Quote:

Originally Posted by littlejodo $View Post$

Find dr/dx if

r = (lnx)/((x^2)lnx^2) + ln(1/x)^3

r' = lnx(x^2 lnx^2)^-1 + 3ln(1/x)

r' = 1/x(x^2 lnx^2)^-1 - lnx(x^2 lnx^2)(2x/x^2) + 3x --> first term I used chain and product rules, but I'm not sure if I carried them out correctly.

then I tried to combine stuff...

r' = 1/(x(x^2lnx^2) - lnx(x^2 lnx^2)(2x/x^2) + 3 --> is this the correct answer? None of the examples I saw in class still had "ln" in the final answer so I wasn't sure if I messed something up... (a high probability scenario).

Thanks!

Your notation is hard to read. Is there your r

$r = \frac{\ln x}{x^2 \ln (x^2)} + \ln \left( \frac{1}{x^3} \right)$

or

$r = \frac{\ln x}{x^2 \left( \ln x \right)^2}+ \left( \ln \frac{1}{x} \right) ^3$

littlejodo · #3 $Old$ January 22nd, 2009, 04:31 PM

so sorry!

Your first term is right in the first example. (ln(x)) / ((x^2)(ln(x^2)))

The second term is correct in the second example. + ln ((1/x)^3)

How do you make yours look like "normal" writing?

Danny · #4 $Old$ January 22nd, 2009, 04:46 PM

Quote:

Originally Posted by littlejodo $View Post$

so sorry!

Your first term is right in the first example. (ln(x)) / ((x^2)(ln(x^2)))

The second term is correct in the second example. + ln ((1/x)^3)

How do you make yours look like "normal" writing?

So it's

$r = \frac{\ln x}{x^2 \ln (x^2)} + \left( \ln \frac{1}{x} \right) ^3$

First, simplify before differentiating. Note $\ln x^r = r \ln x$

so

$r = \frac{\ln x}{2 x^2 \ln x } + \left( - \ln x \right) ^3 = \frac{1}{2} x^{-2} - \left( \ln x \right) ^3$

so

$r' = - \frac{2}{2} x^{-3} - 3 \left( \ln x \right) ^2 \frac{1}{x} = - \frac{1}{x^3} - \frac{3}{x} \left( \ln x \right) ^2$

Making it "look nornal" - it's latex. There is a heading for this on the main page - have a look.