Question #dbaa1

Question

1272 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-28T14:35:02

# (d^2y)/dx^2=6x+2sec^2xtanx.#

#y=x^3+tanx rArr dy/dx=d/dx{x^3+tanx}#

#=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.!