# How do you differentiate f(x)=1/cos(e^arccos(lnx)) using the chain rule?

Mar 30, 2018

$\frac{\textcolor{b l u e}{f ' \left(x\right) = - \tan \left({e}^{\arccos} \left(\ln x\right)\right) \cdot \sec \left({e}^{\arccos} \left(\ln x\right)\right) \cdot {e}^{\arccos} \left(\ln x\right)}}{\textcolor{b l u e}{x \cdot \sqrt{1 - {\ln}^{2} x}}}$

#### Explanation:

$f \left(x\right) = \frac{1}{\cos} \left({e}^{\arccos} \left(\ln x\right)\right)$

$f ' \left(x\right) = \frac{\frac{d}{\mathrm{dx}} \cos \left({e}^{\arccos} \left(\ln x\right)\right)}{\cos} ^ 2 \left({e}^{\arccos} \left(\ln x\right)\right)$

(Multiply and divide by $\cos \left({e}^{\arccos} \left(\ln x\right)\right)$

Applying chain rule,

$\implies \frac{- \sin \left({e}^{\arccos} \left(\ln x\right)\right) \cdot \left(\left(\frac{d}{\mathrm{dx}}\right) \left({e}^{\arccos} \left(\ln x\right)\right)\right)}{\cos} ^ 2 \left({e}^{\arccos} \left(\ln x\right)\right)$

$\implies \frac{- \sin \left({e}^{\arccos} \left(\ln x\right)\right) \cdot {e}^{\arccos} \left(\ln x\right) \cdot \left(\frac{d}{\mathrm{dx}}\right) \arccos \left(\ln x\right)}{\cos} ^ 2 \left({e}^{\arccos} \left(\ln x\right)\right)$

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

=>- (sin (e^arccos(ln x)) * e^arccos(ln x)) /( x * sqrt(1 - ln^2 x)* cos^2 (e^arccos(ln x))

$\implies - \frac{\tan \left({e}^{\arccos} \left(\ln x\right)\right) \cdot \left(\sec \left({e}^{\arccos} \left(\ln x\right)\right) \cdot {e}^{\arccos} \left(\ln x\right)\right)}{x \cdot \sqrt{1 - {\ln}^{2} x}}$