Find the derivative of the function.
g(t) = t2 ln(e2t + 9)
Your questions are so littered with typos that I can not make out what the questions are. Does t2 mean t^2 or 2 t? Do you mean e^(2 t) or t e^2 ?
To find the derivative of the function g(t) = t^2 ln(e^(2t) + 9), we can use the product rule and chain rule of differentiation.
The product rule states that if we have two functions u(t) and v(t), then the derivative of their product is given by d(uv)/dt = u * dv/dt + v * du/dt.
Let's apply the product rule to g(t) = t^2 ln(e^(2t) + 9):
u(t) = t^2
v(t) = ln(e^(2t) + 9)
Now, let's find du(t)/dt and dv(t)/dt.
du(t)/dt = 2t
To find dv(t)/dt, we can use the chain rule. The chain rule states that if we have a composite function y = f(g(t)), then the derivative dy/dt is given by dy/dt = f'(g(t)) * g'(t). In this case, f(u) = ln(u) and g(t) = e^(2t) + 9.
df(u)/du = 1/u (derivative of ln(u))
dg(t)/dt = 2e^(2t) (derivative of e^(2t))
Now, let's find dv(t)/dt using the chain rule:
dv(t)/dt = df(u)/du * dg(t)/dt
= (1/(e^(2t) + 9)) * 2e^(2t)
= 2e^(2t)/(e^(2t) + 9)
Now, applying the product rule:
d(g(t))/dt = u(t) * dv(t)/dt + v(t) * du(t)/dt
= t^2 * (2e^(2t)/(e^(2t) + 9)) + ln(e^(2t) + 9) * 2t
Therefore, the derivative of g(t) = t^2 ln(e^(2t) + 9) is:
d(g(t))/dt = 2t * (1/(e^(2t) + 9)) + 2t^2 * e^(2t)/(e^(2t) + 9)