What is the derivative of #y=2(x^(3)-1)(3x^(2)+1)^4#?

1 Answer
Aug 11, 2015

Answer:

#y^' = 6x(3x^2 + 1) * (11x^3 +x - 8)#

Explanation:

You can differentiate this function by using the product rule and the chain rule.

If you write #y# as

#y = f(x) * g(x)#

then you can say that the derivative of #y# is

#d/dx(y) = 2 * [[d/dx(f(x))] * g(x) + f(x) * d/dx(g(x))]#

In your case, #f(x) = (x^3-1)# and #g(x) = (3x^2+1)^4#.

#y^' = 2 [d/dx(x^3-1) * (3x^2 + 1)^4 + (x^3-1) * d/dx(3x^2+1)^4]#

To determine #d/dx(g(x))#, use the chain rule for

  • #g(u) = u^4#, with #u = 3x^2 + 1#

#d/dx(g(u)) = d/(du)(u^4) * d/dx(u)#

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

#d/dx(3x^2+1) = 4(3x^2+1)^3 * 6x#

Your target derivative will thus be

#y^' = 2 * [3x^2 * (3x^2+1)^4 + (x^3-1) * 24x * (3x^2+1)^3]#

#y^' = 2 * 3x * (3x^2 + 1) * [x * (3x^2 + 1) + (x^3-1) * 8]#

#y^' = 6x(3x^2 + 1) * (3x^3 + x + 8x^3 - 8)#

#y^' = color(green)(6x(3x^2 + 1) * (11x^3 +x - 8))#