Find the derivative of g(x)=(2+(x^2+1)^4)^3?

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Answer 1 · 2018-06-08T08:06:16

#g'(x) = 24x(2+(x^2+1)^4)^2(x^2+1)^3#

#g(x) = (2+(x^2+1)^4))^3#

Apply power and chain rule.

#g'(x) = 3(2+(x^2+1)^4)^2 * d/dx (2+(x^2+1)^4))#

#d/dx (2+(x^2+1)^4)) = 0+4(x^2+1)^3 * d/dx (x^2+1)#

#d/dx (x^2+1) =2x +0#

Combining terms:

#g'(x) = 3(2+(x^2+1)^4)^2 * 4(x^2+1)^3 * 2x#

#= 24x(2+(x^2+1)^4)^2(x^2+1)^3#

Answer 2 · 2018-06-08T08:07:58

Using the chain rule, the derivative of #g(x)=(2+(x^2+1)^4)^3#
is:
#g'(x) = 24x*(2+(x^2+1)^4)^2*(x^2+1)^3#

I hope I'm interpreting the problem correctly:
#g(x)=(2+(x^2+1)^4)^3#

Using the chain rule:
#g'(x) = 3(2+(x^2+1)^4)^2*4(x^2+1)^3*2x#

Simplifying it a little bit:
#g'(x) = 24x*(2+(x^2+1)^4)^2*(x^2+1)^3#