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I'm going over my homework from calc, and am having some trouble with a few questions. It seems as if I'm just not understanding how to solve the problem:

  1. $$f(x) = \sqrt[5]{\ln x}$$

  2. $$f(x) = \sqrt{x}\ln x$$

  3. $$f(x)= \ln \dfrac{({2x+1})^3}{({3x-1})^4}$$

For the first one, I'm not even sure what to do.

As for the second question I ended up with:

$$f'(x) = \frac{1}{2x} \ln x + \frac{1}{x}\sqrt{x}$$

And for the last question I ended up with:

$${(2x+1)^3\over(3x-1)^4} 3 \ln 2 - 4 \ln 3$$

(Sorry for my poor formatting, I wasn't able to use \frac and put the numerator and denominator to the power of x.)

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I don't know what I was reading, but I've updated it to the correct version. There was no power of four, and it was the fifth root not the third. –  user1327636 Mar 6 '13 at 3:02
    
You need to go over the chain rule, and why did you put those weird-looking exponents to those $\,1'$s in (3)? –  DonAntonio Mar 6 '13 at 3:04
    
The post is updated, it was ment to be the entire num./den. to the power as it shows now. –  user1327636 Mar 6 '13 at 3:07
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2 Answers

Hint: ($1$) Use the chain rule and remember $\sqrt[5]{y}=y^{1/5}$. ($2$) Product rule and use the fact that $\sqrt{y}=y^{1/2}$. For ($3$), use the chain rule and quotient rule. As an alternative to the quotient rule, you can use a property of logarithms that says $$\log\left(\frac{x^a}{y^b}\right)=\log(x^a)-\log(y^b)=a\log(x)-b\log(y).$$

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Nice hints (+1 earlier) –  amWhy Mar 6 '13 at 3:32
    
@amWhy: Thanks :) –  Clayton Mar 6 '13 at 3:34
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For the third one, first recall from prerequisite courses that $$f(x)= \ln \dfrac{({2x+1})^3}{({3x-1})^4} = 3\ln(2x+1) - 4\ln(3x-1).$$

Then differentiate, and remember to apply the chain rule as needed.

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My text book says that the answer to that question is: $$\frac{6}{2x+1} - \frac{12}{3x-1}$$ EDIT: I realize how they got there, thank you. –  user1327636 Mar 6 '13 at 3:17
    
Nice, Michael +1. –  amWhy Mar 6 '13 at 3:31
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