Question #dbaa1

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Answer 1 · 2017-01-28T14:35:02

$\frac{d^{2} y}{{d x}^{2}} = 6 x + 2 {sec}^{2} x tan x .$ $(d^2y)/dx^2=6x+2sec^2xtanx.$

$y = x^{3} + tan x \Rightarrow \frac{d y}{d x} = \frac{d}{d x} {x^{3} + tan x}$ $y=x^3+tanx rArr dy/dx=d/dx{x^3+tanx}$

$= \frac{d}{d x} (x^{3}) + \frac{d}{d x} (tan x) = 3 x^{3 - 1} + {sec}^{2} x$ $=d/dx(x^3)+d/dx(tanx)=3x^(3-1)+sec^2x$

$:. dy/dx=3x^2+(secx)^2$

$:. (d^2y)/dx^2=d/dx(dy/dx)=d/dx(3x^2)+d/dx(secx)^2, i.e.,$

$(d^2y)/dx^2=3(2x)+d/dx(t^2), say, where, t=secx...(ast)$

Here, using the Chain Rule,

$d/dx(t^2)={d/dt(t^2)}{d/dx(t)}=(2t){d/dx(secx)}, so,$

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

$d/dx(t^2)=2sec^2xtanx$

$:., by (ast), (d^2y)/dx^2=6x+2sec^2xtanx.$

Enjoy Maths.!