# How do you differentiate f(x)=x/arcsinsqrt(ln(1/x^2) using the chain rule?

Mar 31, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\arcsin \left(\sqrt{- 2 \ln x}\right) + \frac{1}{\sqrt{- 2 \ln x \left(1 + 2 \ln x\right)}}}{\arcsin \left(\sqrt{- 2 \ln x}\right)} ^ 2$

#### Explanation:

Well... you definitely need the chain rule for this one!

I would recommend first splitting this up using the quotient rule.

$y = \frac{u}{v} \to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

In this case $u = x$ and $v = \arcsin \left(\sqrt{\ln \left(\frac{1}{x} ^ 2\right)}\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[x\right] = 1$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[\arcsin \left(\sqrt{\ln \left(\frac{1}{x} ^ 2\right)}\right)\right]$

This is where the chain rule comes in. Let's use the following substitutions:

$a = {x}^{- 2}$
$b = \ln \left(a\right)$
$c = {b}^{\frac{1}{2}}$

The chain rule states that $y = g \left(u\right)$ where $u = f \left(x\right) \to$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

Applying this we get

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[\arcsin \left(\sqrt{\ln \left(\frac{1}{x} ^ 2\right)}\right)\right] =$

$\frac{d}{\mathrm{dc}} \left[\arcsin \left(c\right)\right] \frac{d}{\mathrm{db}} \left[{b}^{\frac{1}{2}}\right] \frac{d}{\mathrm{da}} \left[\ln a\right] \frac{d}{\mathrm{dx}} \left[{x}^{- 2}\right]$

This gives us

$\frac{1}{\sqrt{1 - {c}^{2}}} \times \frac{1}{2} {b}^{- \frac{1}{2}} \times \frac{1}{a} \times \left(- 2 {x}^{- 3}\right)$

Now, putting in the values we get

$\frac{1}{1 - {\left(\sqrt{\ln \left({x}^{- 2}\right)}\right)}^{2}} \times \frac{1}{2} {\left(\ln \left({x}^{- 2}\right)\right)}^{- \frac{1}{2}} \times {\left({x}^{- 2}\right)}^{- 1} \times - 2 {x}^{- 3}$

This can be simplified to

{dv}/dx=-1/{xsqrt{-2lnx(1+2lnx)

Now we have the values we need for $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

Putting those in we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\arcsin \left(\sqrt{- 2 \ln x}\right) + \frac{1}{\sqrt{- 2 \ln x \left(1 + 2 \ln x\right)}}}{\arcsin \left(\sqrt{- 2 \ln x}\right)} ^ 2$