How do you take the derivative of 9000/(1+.3t+.1t^2)?
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
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
To take the derivative of the function f(t) = 9000/(1 + 0.3t + 0.1t^2), you can use the quotient rule:
1. Start with the quotient rule formula:
If you have a function f(t) = g(t)/h(t), then the derivative of f(t) is given by:
f'(t) = (g'(t) * h(t) - g(t) * h'(t)) / (h(t))^2
2. Identify the functions g(t) and h(t):
In our case, g(t) = 9000 and h(t) = 1 + 0.3t + 0.1t^2.
3. Compute the derivatives of g(t) and h(t):
g'(t) = 0 (constant function derivative is always zero)
h'(t) = 0.3 + 0.2t (derivative of 0.3t + 0.1t^2)
4. Plug the values into the formula:
f'(t) = (0 * (1 + 0.3t + 0.1t^2) - 9000 * (0.3 + 0.2t)) / (1 + 0.3t + 0.1t^2)^2
5. Simplify the expression:
f'(t) = -9000 * (0.3 + 0.2t) / (1 + 0.3t + 0.1t^2)^2
So, the derivative of the given function is f'(t) = -9000 * (0.3 + 0.2t) / (1 + 0.3t + 0.1t^2)^2.