How to differentiate?

Use the chain rule twice.
chain rule: D{f(g(x))} = f'(g(x))*g'(x)

You need D{(ln(1+4x))^4}
f(x) = x^4
g(x) = ln(1+4x)
f'(x) = 4x^3
g'(x) = D{ln(1+4x)} need to apply the chain rule to this (let's use h(x) and k(x) instead of f(x) and g(x))
h(x) = ln(x)
k(x) = 1+4x
h'(x) = 1/x
k'(x) = 4

D{(ln(1+4x))^4} = f'(g(x))*g'(x) = f'(g(x))*h'(k(x))*k'(x)
= 4(ln(1+4x))^3 * 1/(1+4x) * 4
= 16(ln(1+4x))^3 / (1+4x)

Thanks for giving me a hand in this!