derivative of t/(9+t)?

Answer 1

Using the quotient rule, we have:

$$\frac{d}{dt} \left(\frac{t}{9+t}\right) = \frac{(9+t)\frac{d}{dt}(t) - t\frac{d}{dt}(9+t)}{(9+t)^2}$$

Simplifying the derivative of the denominator, we have:

$$\frac{d}{dt} (9+t) = 1$$

Simplifying the derivative of the numerator, we have:

$$\frac{d}{dt} (t) = 1$$

Plugging these into the quotient rule formula, we get:

$$\frac{(9+t)(1) - t(1)}{(9+t)^2} = \frac{9}{(9+t)^2}$$

Therefore, the derivative of $t/(9+t)$ is $\boxed{\frac{9}{(9+t)^2}}$

Answer 2

what a lot of illegible mumbo jumbo!
without all the formatting text,

the quotient rule gives
((9+t)(1) - t(1))/(9+t)^2 = 9/(9+t)^2