# Question dbaa1

Jan 28, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 x + 2 {\sec}^{2} x \tan x .$

#### Explanation:

$y = {x}^{3} + \tan x \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left\{{x}^{3} + \tan x\right\}$

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + {\left(\sec x\right)}^{2}$

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right) + \frac{d}{\mathrm{dx}} {\left(\sec x\right)}^{2} , i . e . ,$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 3 \left(2 x\right) + \frac{d}{\mathrm{dx}} \left({t}^{2}\right) , s a y , w h e r e , t = \sec x \ldots \left(\ast\right)$

Here, using the Chain Rule,

$\frac{d}{\mathrm{dx}} \left({t}^{2}\right) = \left\{\frac{d}{\mathrm{dt}} \left({t}^{2}\right)\right\} \left\{\frac{d}{\mathrm{dx}} \left(t\right)\right\} = \left(2 t\right) \left\{\frac{d}{\mathrm{dx}} \left(\sec x\right)\right\} , s o ,$

=(2t)(secxtanx), &, because, t=secx,#

$\frac{d}{\mathrm{dx}} \left({t}^{2}\right) = 2 {\sec}^{2} x \tan x$

$\therefore , b y \left(\ast\right) , \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 x + 2 {\sec}^{2} x \tan x .$

Enjoy Maths.!