What's the derivative of 0,5ln(x³+4)

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y = (1/2)ln(x³ + 4)
use chain rule

y' = (1/2)*1/(x³ + 4) * 3x²

y' = 3x²/(2x³ + 8)

y = 1/2 * ln(x^3 + 4)
Put the 1/2 on the outside and just focus on d/dx (ln(x^3 + 4))
By the chain rule, we need to take the derivative of the inside then the derivative of the whole thing:
The derivative of the inside is 3x^2.
The derivative of ln(u) = 1/u
So it is 3x^2 * 1/(x^3 + 4) and then multiply the 1/2 we have outside the front:
dy/dx = 1/2 * 3x^2 * 1/(x^3 + 4)
dy/dx = 3x^2 / (2(x^3 + 4))

d[0,5ln(x³+4)]\dx = 0,5d[ln(x³+4)]\dx = 0,5[ (1/(x³+4)) * 3x²]
this is an application of the chain rule, the derivative of a composition of functions
take the derivative of the outside, multiply by the derivative of the inside

1/2*ln(x³+4)*3x²