Help finding the derivative?

Question

Find the derivative: y=t^2/3*logbase3*Sqroot(t^3+1)

Answer 1

OK let's see if I understand what you've written.

y = t^(2/3) * log3 (t^3 + 1)^(1/2)
= 1/2 * t^(2/3) * log3 (t^3 + 1)
= 1/(2*ln3) * t^(2/3) * ln(t^3 + 1)

dy/dt =
1/(2*ln3)*[(2/3)*t^(-1/3)* ln(t^3 + 1) + t^(2/3) * 3t^2 / (t^3 + 1)]

I'm sure you could simplify it from there.

Answer 2

Your notation isn't all that clear. I'll assume you mean
y = t^(2/3) log_3 (√(t^3+1))
= t^(2/3) ln (√(t^3+1)) / ln 3
where I'm using log_3 for log base 3, to save space.
This is just product rule and chain rule (a few times for chain rule):
y' = (2/3) t^(-1/3) log_3 (√(t^3+1)) + t^(2/3) . [1 / √(t^3 + 1)] . [(1/2) (t^3 + 1)^(-1/2)] . (3t^2) / ln 3
= (1/3) t^(-1/3) log_3 (t^3+1) + 3 t^(8/3) / [2 (t^3 + 1) ln 3]

noting that log_3 (√(t^3+1)) = log_3 (t^3+1) / 2 by the laws of logs.