Take the derivative of and I'm just confused about where I stop for the first one, how much do I have to simplify. And for the second one, I don't know how to start it. Thanks in advance!
For the second one, I know it looks scary, but break it up using the log rules and it ain't so bad.
$\displaystyle ln\left(\frac{x^{5}e^{4x}\sqrt{3x+1}}{1-x^{2}}\right)$
=$\displaystyle \left[ln(x^{5})+ln(e^{4x})+ln(\sqrt{3x+1})\right]-ln(1-x^{2})$
=$\displaystyle 5ln(x)+4x+\frac{1}{2}ln(3x+1)-ln(1-x^{2})$
Now, you can differentiate, can't you?.
$\displaystyle f(x)= x \cdot ln(x + 3) - ln(9)$
$\displaystyle f'(x) = ln(x + 3) + \frac{x}{x + 3}$
I would leave it like this.
For the second problem, the first thing I would do would be to use the property of logarithms to split this up a bit:
$\displaystyle ln \left ( \frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2} \right )$
= $\displaystyle ln(x^5) + ln(e^{4x}) + ln(\sqrt{3x+1}) - ln(1-x^2)$
= $\displaystyle 5 ln(x) + 4x + \frac{1}{2} \cdot ln(3x + 1) - ln(1 - x^2)$
Then
$\displaystyle \frac{d}{dx}ln \left ( \frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2} \right )$
= $\displaystyle \frac{5}{x} + 4 + \frac{1}{2} \cdot \frac{3}{3x+1} - \frac{1}{1 - x^2} \cdot (-2x)$
and then simplify a bit.
The brute force approach also works. It involves the chain rule, the quotient rule, the product rule, and a LOT of cancellations:
$\displaystyle \frac{d}{dx}ln \left ( \frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2} \right )$
= $\displaystyle \frac{1}{\frac{x^5e^{4x}\sqrt{3x+1}}{1-x^2}}$$\displaystyle \cdot ( \frac{5x^4e^{4x}\sqrt{3x+1}}{(1-x^2)}$ $\displaystyle +\frac{x^5 \cdot 4e^{4x} \sqrt{3x+1}}{(1-x^2)}$ $\displaystyle + \frac{x^5e^{4x} \cdot \frac{3}{2\sqrt{3x+1}}}{(1 - x^2)}$ $\displaystyle - \frac{x^5e^{4x}\sqrt{3x+1} \cdot (-2x)}{(1-x^2)^2} )$
(Sorry. I had to split the second factor up into terms to make the LaTeX work.)
= $\displaystyle \frac{5}{x} + 4 + \frac{3}{2(3x + 1)} + \frac{2x}{1-x^2}$
just like before. (I highly recommend the first method, but the second one is *great* for practice! )
-Dan