Chain rule?

Question

What is the chain rule of calculus? I do not really understand the process. Use f(x) = (x^2+1)^3 as an example please

Answer 1

f(x) = (x²+1)³

Using it to find the derivative...
First drop the 3 down in front of the brackets and then the brackets will be raised to n-1, which is 3-1 = 2.
Giving you: [3(x² +1)²]

f(x)´ = 3(x²+1)² * (2x + 0)

Then it's multiplied by what's the derivative inside the brackets. Using the same method and only what's inside the original brackets this time, drop the 2 down in front of the x, then it'll be raised to n-1, which is 2-1 = 1. Only x.
Giving you: [2x + 0]

The derivative of 1 (or any constant) is zero. So, that isn't present in the equation any longer. [2x + 0] is just 2x
f(x)´ = 3(x²+1)² * (2x)

Now, you can just continue it out as in normal algebra.
f(x)´ = 3(x²+1)(x²+1) * (2x)
f(x)´ = 3(x^4 + x² + x² + 1) * (2x)
f(x)´ = 3(x^4 + 2x² + 1) * (2x)
f(x)´ = (3x^4 + 6x² + 3) * (2x)
f(x)´ = 6x^5 + 12x³ + 6x <----------Answer.

Answer 2

d(u³)/du = 3u²
d(x² + 1)/dx = 2x
so see x² + 1 as u,
d[(x²+1)³]dx = du³/du • du/dx =
3(x²+1)²(2x) =
6x(x^4 + 2x² + 1) =
6x^5 + 12x³ + 6x