I don't understand chain rule for differentiation especially when combined with more complex functions.

The chain rule is a method that allows you to differentiate a function of a function, which can be of the form y = f[g(x)]. That is, g is a function of x and f is a function of g, which in turn is a function of x. An example could be y = (5x²+1)^1/2. In this case g(x) = 5x²+1 and f(x) = x^1/2, so when we write f[g(x)], g(x) takes the role of the variable x. Chain rule states that if y = f[g(x)], then dy/dx = f'[g(x)]g'(x). 
Applying this definition:

f'(x) = 0.5*x^-1/2

g'(x) = 10x

Combine the results and replace x with g(x) in f'(x), then:

dy/dx = f'[g(x)]g'(x) = 0.5(5x²+1)^-1/2(10x) = 5x(5x²+1)^-1/2 which is the final result. 

The way I like to think of this method is to identify the main function which I know how to differentiate without applying the chain rule (in this case y is of the form xⁿ). Whatever is within this function (probably another function - in this case 5x²+1) replace it with x. Then differentiate like the inner function is simply the variable x (dy/dx = 0.5(5x²+1)^-1/2) and finally multiply the result with the derivative of the variable x, which is now a function .((5x²+1)' = 10x -> dy/dx = 0.5(5x²+1)^-1/2*(10x) = 5x(5x²+1)^-1/2 ).  

Here is an example of a more complex number: 

f(x) = x²(3x-1)^-1/2

This example requires the application of the product rule and the chain rule. 

Product rule states that: (uv)' = u'v + uv'

In this example u = x² and v = (3x-1)^1/2

u' = 2x and v' = 0.5(3x-1)^-1/2*3 = 1.5(3x-1)^-1/2  (by applying the chain rule - think of 3x-1 simply as x and apply the above procedure)

Then, 

f'(x) = (2x) (3x-1)^1/2 + (x²)(1.5(3x-1)^-1/2 ​)

which can then be simplefied to [x(15x-4)]/[2(3x-1)^1/2]