# Thread: If $$y = In \left[ \frac{x(x^2 + 1)}{\sqrt{x^3 - 1}} \right]$$

1. ## If $$y = In \left[ \frac{x(x^2 + 1)}{\sqrt{x^3 - 1}} \right]$$

$y = In \left[ \frac{x(x^2 + 1)}{\sqrt{x^3 - 1}} \right]$ find y'

I began this by
letting $u=\frac{x(x^2+1)}{\sqrt{x^3 - 1}}$

$y' = \frac{1}{u}.u'$

$\frac{3x}{x^2 + 1} + \frac{1}{x(x^2+1)} - \frac{3x^2}{2(x^3 - 1)}$

which is wrong

2. What are you trying to do? Differentiate this function?

$\displaystyle y = \ln{\left[\frac{x(x^2 + 1)}{\sqrt{x^3 - 1}}\right]}$

$\displaystyle e^y = (x^3 + x)(x^3 - 1)^{-\frac{1}{2}}$

$\displaystyle \frac{d}{dx}\left(e^y\right) = \frac{d}{dx}\left[(x^3 + x)(x^3 - 1)^{-\frac{1}{2}}\right]$

$\displaystyle \frac{d}{dy}\left(e^y\right)\,\frac{dy}{dx} = (x^3 + x)\,\frac{d}{dx}\left[(x^3 - 1)^{-\frac{1}{2}}\right] + (x^3 - 1)^{-\frac{1}{2}}\,\frac{d}{dx}(x^3 + x)$.

Go from here...

3. Originally Posted by Prove It
What are you trying to do? Differentiate this function?

$\displaystyle y = \ln{\left[\frac{x(x^2 + 1)}{\sqrt{x^3 - 1}}\right]}$

$\displaystyle e^y = (x^3 + x)(x^3 - 1)^{-\frac{1}{2}}$

$\displaystyle \frac{d}{dx}\left(e^y\right) = \frac{d}{dx}\left[(x^3 + x)(x^3 - 1)^{-\frac{1}{2}}\right]$

$\displaystyle \frac{d}{dy}\left(e^y\right)\,\frac{dy}{dx} = (x^3 + x)\,\frac{d}{dx}\left[(x^3 - 1)^{-\frac{1}{2}}\right] + (x^3 - 1)^{-\frac{1}{2}}\,\frac{d}{dx}(x^3 + x)$.

Go from here...
thank you, yes, I am trying to differentiate it

4. $\displaystyle \frac{d}{dy}\left(e^y\right)\,\frac{dy}{dx} = (x^3 + x)\,\frac{d}{dx}\left[(x^3 - 1)^{-\frac{1}{2}}\right] + (x^3 - 1)^{-\frac{1}{2}}\,\frac{d}{dx}(x^3 + x)$.

I am a bit lost and rusty, I could do this a month back but now I've forgotten what to do.

the right hand side looks like this
$\displaystyle = (x^3 + x)\,(-\frac{1}{2}(x^3 - 1)^{-\frac{3}{2}} 3x^2) + (x^3 - 1)^{-\frac{1}{2}}\,(3x^2 + 1)$

what do I do now