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

2 Answers
Jun 8, 2018

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

Explanation:

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

Jun 8, 2018

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

Explanation:

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