How do you take the derivative of 9000/(1+.3t+.1t^2)?

Answer 1

Now I am going to have to use the table formulas because this is getting to be too much to do from scratch. You can take the constant 9000 outside immediately

The basic rule for this is
d/dx( u/v) = [v du/dx - u dv/dx] / v^2
or the way to remember it is
[bottom derivative of top - top derivative of bottom]/bottom squared

here we have for
9000 d/dt [1/(1 +.3 t + .1t^2)]=

9000 [ 0 - 1 (d/dt (1+.3t+.1t^2) )]/(1+.3t+.1t^2)^2

which is
9000[ -.3 +.2t ] /(1 +.3t +.1t^2)^2

I can not do a lot more with this. I hope this is part of some problem where you have a value for t

Answer 2

write it as 9000(1+.3t+.1t^2)^(-1)

then derivative = -9000(1+.3t+.1t^2)^(-2)(.3+.2t)
=-9000(.3+.2t)/(1+.3t+.1t^2)^2