Differentiation of logarithmic functions

Question

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.)

Answer 1

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 $$\log\left(\frac{x^a}{y^b}\right)=\log(x^a)-\log(y^b)=a\log(x)-b\log(y).$$

Answer 2

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.