# How do you differentiate f(x)=1/cossqrt(lnx) using the chain rule?

$f ' \left(x\right) = \frac{\setminus \sec \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \tan \left(\setminus \sqrt{\setminus} \ln x\right)}{2 x \setminus \sqrt{\setminus} \ln x}$

#### Explanation:

Given that

$f \left(x\right) = \frac{1}{\setminus \cos \setminus \sqrt{\setminus \ln x}}$

$f \left(x\right) = \setminus \sec \setminus \sqrt{\setminus \ln x}$

Differentiating w.r.t. $x$ using chain rule as follows

$f \left(x\right) = \setminus \sec \setminus \sqrt{\setminus \ln x}$

$= \frac{d}{\mathrm{dx}} \left(\setminus \sec \left(\setminus \sqrt{\setminus \ln x}\right)\right)$

$= \setminus \sec \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \tan \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \frac{d}{\mathrm{dx}} \left(\setminus \sqrt{\setminus} \ln x\right)$

$= \setminus \sec \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \tan \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \frac{1}{2 \setminus \sqrt{\ln} x} \frac{d}{\mathrm{dx}} \setminus \ln x$

$= \setminus \sec \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \tan \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \frac{1}{2 \setminus \sqrt{\ln} x} \frac{1}{x}$

$= \frac{\setminus \sec \left(\setminus \sqrt{\setminus} \ln x\right) \setminus \tan \left(\setminus \sqrt{\setminus} \ln x\right)}{2 x \setminus \sqrt{\setminus} \ln x}$

Jul 18, 2018

$f ' \left(x\right) = \frac{1}{2 x \sqrt{\ln x}} \sec \left(\sqrt{\ln x}\right) \tan \left(\sqrt{\ln x}\right)$

#### Explanation:

Here ,

$f \left(x\right) = y = \frac{1}{\cos} \sqrt{\ln} x$

Let ,

$y = \frac{1}{u}$ , $u = \cos v$ , $v = \sqrt{w}$ , $w = \ln x$

So,

$\frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{u} ^ 2$ , $\frac{\mathrm{du}}{\mathrm{dv}} = - \sin v$ ,  (dv)/(dw)=1/(2sqrtw)and(dw)/(dx)=1/x

Using Chain Rule :

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dw}} \frac{\mathrm{dw}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \frac{1}{u} ^ 2\right) \left(- \sin v\right) \left(\frac{1}{2 \sqrt{w}}\right) \left(\frac{1}{x}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} ^ 2 v \cdot \sin v \left(\frac{1}{2 \sqrt{w}}\right) \left(\frac{1}{x}\right) \to \left[\because u = \cos v\right]$

=>(dy)/(dx)=sin(sqrtw)/cos^2(sqrtw)*1/(2sqrtw) (1/x)to[becausev=sqrtw]

=>(dy)/(dx)=sin(sqrt(lnx))/cos^2(sqrt(lnx))1/(2sqrt(lnx)) (1/x)to[becausew=lnx]

Hence,

$f ' \left(x\right) = \frac{1}{2 x \sqrt{\ln x}} \sin \frac{\sqrt{\ln} x}{\cos} ^ 2 \left(\sqrt{\ln x}\right) \mathmr{and}$

$f ' \left(x\right) = \frac{1}{2 x \sqrt{\ln x}} \sec \left(\sqrt{\ln x}\right) \tan \left(\sqrt{\ln x}\right)$