Differentiate f(x) = 4x(x^2 + 1)^3

Using the product rule and chain rule I can get it to:
f'(x) = 4x * 3(x^2 + 1)^2 * 2x + 4(x^2 + 1)^3
but then my solution guide says:
f'(x) = 4(x^2 + 1)^2 * (7x^2 + 1) as the final step.
How does one arrive at this solution? (or is this a misprint) THANKS!

No need to fear yet, all that has taken place there is some simple simplification.

We can rewrite this derivative as:

4x * 3(x^2+1)^2 * 2x + 4(x^2 + 1)(x^2 + 1)^2

Now just do a little more simplification and that answer is bound to show up.

(24x^2)(x^2 + 1)^2 + 4(x^2 + 1)(x^2 + 1)^2
4(6x^2)(x^2 + 1)^2 + 4(x^2 + 1)(x^2 + 1)^2
4(x^2 + 1)^2(6x^2 + x^2 + 1) = 4(x^2 + 1)^2 * (7x^2 + 1)

Hope this helped!

Ok, we have to remember theres a chain rule + product rule
f`g + g`f

f`(x) = 4(x^2 + 1)^3 + 3(x^2 + 1)^2 (2x)(4x)
=4(x^2 + 1)^3 +24x(x^2 + 1)^2
Then factor
=4(x^2 + 1)^3 +24x(x^2 + 1)^2
=4(x^2 + 1)^2 [x^2 + 6x]

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I got the same answer as you:
f'(x) = 4x[3(x^2+1)^2(2x)] + [(x^2+1)^3](4)
f'(x) = (24x^2)(x^2+1)^2 + 4(x^2+1)^3

It must be a misprint.

:)

Using product rule along with chain rule:

f '(x) = 4x*(6x)(x² + 1)² + 4*(x² + 1)^3

f '(x) = (x² + 1)²[28x² + 4]

4x * 3(x^2 + 1)^2 * 2x + 4(x^2 + 1)^3
= 4(x^2 + 1)^2 (6x^2 + x^2 + 1)
= 4(x^2 + 1)^2 * (7x^2 + 1)