What is the second derivative of # (x^2-1)^3#?

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Answer 1 · 2017-01-07T19:08:16

#(d^2)/(dx^2) (x^2-1)^3=6(x^2-1)(5x^2-1)#

Let's calculate the first derivative using the chain rule:

#d/(dx) (x^2-1)^3= 3(x^2-1)^2*d/(dx) (x^2-1) = 6x(x^2-1)^2#

Now using the product rule:

#(d^2)/(dx^2) (x^2-1)^3 = d/(dx) [ 6x(x^2-1)^2] = 6(x^2-1)^2+24x^2(x^2-1)=6(x^2-1)(x^2-1+4x^2)=6(x^2-1)(5x^2-1)#

Answer 2 · 2017-01-07T19:19:11

# (d^2y)/dx^2=6(5x^4-6x^2+1).#

Let #y=(x^2-1)^3#

#:. dy/dx=3(x^2-1)^2d/dx(x^2-1)=6x(x^2-1)^2#

#:. (d^2y)/dx^2=d/dx(dy/dx)=d/dx{6x(x^2-1)^2}#

#=6d/dx{x(x^2-1)^2}#

#=6[x{d/dx(x^2-1)^2}+(x^2-1)^2{d/dxx}]...["Product Rule]"#

#=6[x{2(x^2-1)d/dx(x^2-1)}+(x^2-1)^2(1)]#

#=6[x{4x(x^2-1)}+(x^2-1)^2]#

#=6(x^2-1){4x^2+(x^2-1)}=6(x^2-1)(5x^2-1)#

#:. (d^2y)/dx^2=6(5x^4-6x^2+1).#

Alternatively, we can expand

#(x^2-1)^3" as "(x^2)^3-1^3-3x^2(x^2-1)#

and get, #y=x^6-1-3x^4+3x^2#

#:. dy/dx=6x^5-12x^3+6x#

#:. (d^2y)/dx^2=30x^4-36x^2+6=6(5x^4-6x^2+1),# as before!

Enjoy Maths.!