How do you differentiate f(x)=sec(1/sqrt(3x^2-4) )  using the chain rule?

Jun 26, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 x}{3 {x}^{2} - 4} ^ \left(\frac{3}{2}\right) \sec \left(\frac{1}{\sqrt{3 {x}^{2} - 4}}\right) \tan \left(\frac{1}{\sqrt{3 {x}^{2} - 4}}\right)$

Explanation:

Let ,

color(red)(y=secu  , color(red)(u=1/v and v=sqrt(3x^2-4

$\implies \frac{\mathrm{dy}}{\mathrm{du}} = \sec u \tan u ,$ $\frac{\mathrm{du}}{\mathrm{dv}} = - \frac{1}{v} ^ 2 \mathmr{and}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{2 \sqrt{3 {x}^{2} - 4}} \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 4\right) = \frac{6 x}{2 \sqrt{3 {x}^{2} - 4}} = \frac{3 x}{\sqrt{3 {x}^{2} - 4}}$

Using chain rule:

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dv) .(dv)/(dx)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec u \tan u \left(- \frac{1}{v} ^ 2\right) \frac{3 x}{\sqrt{3 {x}^{2} - 4}} , w h e r e , u = \frac{1}{v}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sec \left(\frac{1}{v}\right) \tan \left(\frac{1}{v}\right) \left(- \frac{1}{v} ^ 2\right) \frac{3 x}{\sqrt{3 {x}^{2} - 4}}$

Putting , $v = \sqrt{3 {x}^{2} - 4}$

$\frac{\mathrm{dy}}{\mathrm{dx}}$=$- \sec \left(\frac{1}{\sqrt{3 {x}^{2} - 4}}\right) \tan \left(\frac{1}{\sqrt{3 {x}^{2} - 4}}\right) \frac{1}{3 {x}^{2} - 4} \cdot \frac{3 x}{\sqrt{3 {x}^{2} - 4}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 x}{3 {x}^{2} - 4} ^ \left(\frac{3}{2}\right) \sec \left(\frac{1}{\sqrt{3 {x}^{2} - 4}}\right) \tan \left(\frac{1}{\sqrt{3 {x}^{2} - 4}}\right)$