The derivative of $ \ln\left(\frac{x+2}{x^3-1}\right)$ I know it is a simple question, but what would be the next few steps in this equation to find the derivative?
$$f(x)= \ln\left(\frac{x+2}{x^3-1}\right)$$
 A: Let:
$$h(x) = \ln(x) \hspace{.3cm} \text{and} \hspace{.3cm} g(x) = \frac{x + 2}{x^3 - 1}$$
so that:
$$f(x) = h(g(x))$$
Now using the chain rule:
$$f'(x) = h'(g(x))g'(x)$$
A: For $$f(x)= \ln\left(\frac{x+2}{x^3-1}\right)$$
Start by re-writing the logarithm as
$$\frac{d}{dx}\left(\ln\left(\frac{x+2}{x^3-1}\right)\right)=\frac{d}{dx}\left(\ln({x+2})-\ln({x^3-1})\right)$$$$=\frac{d}{dx}\left(\ln({x+2})\right)-\frac{d}{dx}\left(\ln({x^3-1})\right)= \color{blue}{\frac{1}{x+2}-\frac{3x^2}{x^3-1}}$$
A: It would also be advantageous here to apply a property of logarithms, specifically:
$$
\ln(a/b)=\ln(a)-\ln(b).
$$
Applying this to your given equation will considerably simplify your derivative, as it will eliminate your need to use the quotient rule as part of your chain rule calculation.
A: I would NOT use the chain rule before doing this:
\begin{align}
f(x) = \ln\frac{x+2}{x^3-1} & = \ln (x+2) - \ln(x^3-1) \\[10pt]
& = \ln(x+2) - \ln((x-1)(x^2+x+1)) \\[10pt]
& = \ln(x+2) - \ln(x+1) - \ln(x^2+x+1).
\end{align}
A: I suppose that you know that the derivative of $y=\ln(x)$ is $y'=\dfrac{1}{x}$, and you know the chain rule that gives the derivative of a composite function:
$y=f(g(x)) \rightarrow y'=f'(g(x))g'(x)$.
Use these  and you have:
$$y=\ln\frac{x+2}{x^3-1} \rightarrow y'=\frac{x^3-1}{x+2} \left(\frac{x+2}{x^3-1} \right)'$$
Now use the derivative of a fraction.... and you find the result. 
A: Look up the chain rule on youtube. There is a lot of songs ect that will make this easy as cake. 
A: $$f'_{(x)}=\frac{1}{\frac{x+2}{x^3-1}}\left (\frac{(x^3+1)-(x+2)3x^2}{(x^3-1)^2}\right)$$
$$f'_{(x)}=\frac{-(2x^3+6x^2-1)}{(x+2)(x^3-1)}$$
