# How to differentiate y= ln^3(cos^2 √(1-x)?

Aug 11, 2015

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

#### Explanation:

You have to use the Chain Rule five times:

If $y = f \left(x\right) = {\ln}^{3} \left({\cos}^{2} \left(\sqrt{1 - x}\right)\right)$, then

$f ' \left(x\right) = 3 {\ln}^{2} \left({\cos}^{2} \left(\sqrt{1 - x}\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln \left({\cos}^{2} \left(\sqrt{1 - x}\right)\right)\right)$

$= 3 {\ln}^{2} \left({\cos}^{2} \left(\sqrt{1 - x}\right)\right) \cdot \frac{1}{{\cos}^{2} \left(\sqrt{1 - x}\right)} \cdot \frac{d}{\mathrm{dx}} \left({\cos}^{2} \left(\sqrt{1 - x}\right)\right)$

$= \frac{3 {\ln}^{2} \left({\cos}^{2} \left(1 - x\right)\right)}{{\cos}^{2} \left(\sqrt{1 - x}\right)} \cdot 2 \cos \left(\sqrt{1 - x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\cos \left(\sqrt{1 - x}\right)\right)$

$= \frac{6 {\ln}^{2} \left({\cos}^{2} \left(1 - x\right)\right)}{\cos \left(\sqrt{1 - x}\right)} \cdot \left(- \sin \left(\sqrt{1 - x}\right)\right) \cdot \frac{d}{\mathrm{dx}} \left({\left(1 - x\right)}^{\frac{1}{2}}\right)$

$= - 6 {\ln}^{2} \left({\cos}^{2} \left(1 - x\right)\right) \tan \left(\sqrt{1 - x}\right) \cdot \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 - x\right)$

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