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Math Help - Calc2 - Dx with ln

  1. #1
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    Calc2 - Dx with ln

    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!
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  2. #2
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    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
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  3. #3
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    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?
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  4. #4
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    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.
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