G (t) = t^2e^-t + ((ln (t)) ^2)

Find the derivative of the following via implicit differentiation:d/dt (G (t)) = d/dt (t^2/e^t + log^2 (t))

The derivative of G (t) is G' (t) : G' (t) = d/dt (t^2/e^t + log^2 (t))

Differentiate the sum term by term:G' (t) = d/dt (t^2/e^t) + d/dt (log^2 (t))

Use the product rule, d/dt (u v) = v (du) / (dt) + u (dv) / (dt), where u = e^ (-t) and v = t^2:G' (t) = d/dt (log^2 (t)) + t^2 d/dt (e^ (-t)) + (d/dt (t^2)) / e^t

Simplify the expression:G' (t) = t^2 (d/dt (e^ (-t))) + (d/dt (t^2)) / e^t + d/dt (log^2 (t))

Using the chain rule, d/dt (e^ (-t)) = (d e^u) / (du) (du) / (dt), where u = - t and (d) / (du) (e^u) = e^u:G' (t) = (d/dt (t^2)) / e^t + d/dt (log^2 (t)) + (d/dt (-t)) / e^t t^2

Factor out constants:G' (t) = (d/dt (t^2)) / e^t + d/dt (log^2 (t)) + (-d/dt (t) t^2) / e^t

The derivative of t is 1:G' (t) = (d/dt (t^2)) / e^t + d/dt (log^2 (t)) - (1 t^2) / e^t

Use the power rule, d/dt (t^n) = n t^ (n - 1), where n = 2: d/dt (t^2) = 2 t:G' (t) = - t^2/e^t + d/dt (log^2 (t)) + (2 t) / e^t

Using the chain rule, d/dt (log^2 (t)) = (du^2) / (du) (du) / (dt), where u = log (t) and (d) / (du) (u^2) = 2 u:G' (t) = (2 t) / e^t - t^2/e^t + 2 log (t) d/dt (log (t))

The derivative of log (t) is 1/t:Answer: G' (t) = (2 t) / e^t - t^2/e^t + 1/t 2 log (t)