What are the first and second derivatives of a Surge Function? (product rule)

y= 8te^-0.45t

I need these 2 answers to calculate the pt of inflection (where 2nd deriv. =0) and the local maximum (where 1st deriv. =0).

All help is apreciated! =)

Use product rule d(uv)/dx=udv/dx+vdu/dx
u=8t
v=exp(-0.45t)
du/dx=8
dv/dx=-0.45exp(-0.45t)
y'=-3.6texp(-0.45t)+8exp(-0.45t)
Use the same method for the second derivative
y''=1.62texp(-0.45t)-7.2exp(-0.45t)

You need to use the product rule AND the chain rule...

First Derivative:
(8e^-0.45t) + (8te^-0.45t)(-0.45)
= 8e^-0.45t – 3.6te^-0.45t

Second Derivative:
(e^-0.45t)(-0.45) - [3.6e^-0.45t + 3.6t(e^-0.45t)(-0.45)]
= (1.62te^-0.45t) - (4.05e^-0.45t)

y= (8t)(e^-0.45t)
1st derivative: 8e^(-0.45t)-3.6te^(-0.45t)
2nd derivative: -3.6e^(-0.45t)-3.6e^(-0.45t)+1.62te^(-0.…
-7.2e^(-0.45t)+1.62te^(-0.45t)