How do you find the second derivative of #y = x^4 tan x#?

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Answer 1 · 2015-02-21T15:34:17

The answer is:

#y''=2x^2(4x+6tanx+4xtan^2x+x^2tan^2x+x^2tan^3x)#.

With the pruduct rule that says:

#y=f(x)*g(x)rArry'=f'(x)*g(x)+f(x)g'(x)#

Than:

#y'=4x^3tanx+x^4(1+tan^2x)#

or

#y'=4x^3tanx+x^4*1/cos^2x#.

From the first:

#y''=12x^2tanx+4x^3(1+tan^2x)+4x^3(1+tan^2x)+x^4*2tanx*(1+tan^2x)=#

#=2x^2(6tanx+2x(1+tan^2x)+2x(1+tan^2x)+x^2tanx(1+tan^2x))=#

#=2x^2(6tanx+2x+2xtan^2x+2x+2xtan^2x+x^2tanx+x^2tan^3x)=#

#=2x^2(4x+6tanx+4xtan^2x+x^2tan^2x+x^2tan^3x)#.