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$f(x)=ln(\sqrt {\frac {6x+3}{3x-9}})$

Find $f'(x)$.

I've tried the method involving distributing the natural log and making it $(1/2)\ln(6x+3)-(1/2)\ln(3x-9)$.

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  • $\begingroup$ why didnt you proceed further? $\endgroup$
    – Jasser
    Oct 11, 2014 at 16:36
  • $\begingroup$ I did. I just got the wrong answer, so I started back from there $\endgroup$
    – Sherbs
    Oct 11, 2014 at 16:39
  • $\begingroup$ Yes, both the numerator and the denominator are under the square root $\endgroup$
    – Sherbs
    Oct 11, 2014 at 16:44
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    $\begingroup$ I changee the title, since "logarithmic differentiation" means something different from what you are doing here. $\endgroup$ Oct 11, 2014 at 16:45

1 Answer 1

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$$\frac {d(logG(x))}{dx}= \frac 1{logG(x)} \frac {d(G(x))}{dx}$$

$$df/dx=\frac {\frac {6}{6x+3} - \frac {3}{3x-9}}2$$

which can be simplified to

$$\frac {-7}{2(2x+1)(x-3)}$$

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  • $\begingroup$ Thank you. I ended up getting that answer myself. I got the problem wrong because I didn't further simplify the fraction from 1/(2x+1)-1/(2x-6) $\endgroup$
    – Sherbs
    Oct 11, 2014 at 16:56
  • $\begingroup$ I am happy that it helped :). But hey from where did you get 2x-6 ... $\endgroup$
    – Jasser
    Oct 11, 2014 at 17:00
  • $\begingroup$ I distributed the ln to get (1/2)ln(6x+3)-(1/2)ln(3x-9), then getting (1/2)(6/6x+3)-(1/2)(3/3x-9). I factored the two terms getting 1/(2x+1)-1/2(x-3). I got (2x-6) from multiplying the 2 with x-3 $\endgroup$
    – Sherbs
    Oct 11, 2014 at 17:06
  • $\begingroup$ Oh yes, silly me 2(x-3)=2x-6. Enjoy learning. $\endgroup$
    – Jasser
    Oct 11, 2014 at 17:11

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