Logarithmic differentiation - need help please Determine the following by using logarithmic differentiation.
$$y=\frac{\sqrt{x}(x^2-1)^5}{\cos(x))}$$
My answer is up to:
Take the ln on both sides so:
$\ln(y) = (1/2)\ln(x) + 5\ln(x^2-1) - \ln(\cos(x))$
Am I doing it right though?
 A: you must differetiate both sides with respect to $x$
$$\frac{y'}{y}=\frac{1}{2x}+\frac{5}{x^2-1}2x+\frac{\sin(x)}{\cos(x)}$$
after simplification we get
$$y'=\frac{\sqrt{x}(x^2-1)^5}{\cos(x)}\left(\frac{1}{2x}+\frac{10x}{x^2-1}+\tan(x)\right)$$
A: Lets apply the following rule:

If $u$ is a positive function of $x$ then $(\ln u)'=\frac{u'}{u}$.

Well, as what I have said, lets apply term by term differentiation by using the above rule to get
$$\frac{y'}{y}=\frac{1}{2}\frac{1}{x}+5\frac{2x}{x^2-1}-\frac{-\sin x}{\cos x}.$$ Thus, we get
$$y'=y\bigg[\frac{1}{2x}+\frac{10x}{x^2-1}+\frac{\sin x}{\cos x}\bigg],$$ then plug in the value of $y$ to get
$$y'=\frac{\sqrt{x}(x^2-1)^5}{\cos x}\bigg[\frac{1}{2x}+\frac{10x}{x^2-1}+\tan x\bigg]$$
A: You have everything right so far.
$\ln(y) = (1/2)\ln(x) + 5\ln(x^2-1) - \ln(\cos(x))$
Now remember, that the derivative of $\ln(y) = \frac{1}{y}\times y'$
This is because $y$ is a function, whereas $x$ is just a variable. 
The derivative of $\ln(x)$ in this case is just $\frac{1}{x}$
Also remember that if you have something like $\ln(x^2 + 1)$, that's chain rule.
It's the derivative of $\frac{1}{x^2-1}\times\frac{d}{dx}(x^2-1)$
Differentiating both sides, we get:
$$\frac{1}{y}\times y' = (\frac{1}{2}\cdot\frac{1}{x}) + (5\cdot\frac{1}{x^2-1}\cdot2x) - (\frac{1}{\cos(x)}\cdot-\sin(x))$$
$$y' = y(\frac{1}{2x} + \frac{10x}{x^2-1}) + \tan(x)$$
