# Thread: Derivatives problem

1. ## Derivatives problem

Hello, back again with another derivatives question:

A grocery store determines that after t hours on the job, a new cashier can ring up N(t) = 20 - (30
÷
√(9 + t^2)

items per minute. Find N' (t), the rate at which the cashier's productivity is changing.

I've solved the derivative to

(9 + t^2)^(1/2) - 30 t
÷
(9+t^2)^(3/2)

But the correct answer should be simply

30 t
÷
(9 + t^2)^(3/2)

I have no idea how the (9 + t^2)^(1/2) in the numerator vanishes, though I've tried re-doing it several times. I'd greatly appreciate it if anyone could show me the steps to simplifying this derivative. Thank you in advance!

2. Originally Posted by starswept
Hello, back again with another derivatives question:

A grocery store determines that after t hours on the job, a new cashier can ring up N(t) = 20 - (30
÷
√(9 + t^2)

items per minute. Find N' (t), the rate at which the cashier's productivity is changing.

I've solved the derivative to

(9 + t^2)^(1/2) - 30 t
÷
(9+t^2)^(3/2)

But the correct answer should be simply

30 t
÷
(9 + t^2)^(3/2)

I have no idea how the (9 + t^2)^(1/2) in the numerator vanishes, though I've tried re-doing it several times. I'd greatly appreciate it if anyone could show me the steps to simplifying this derivative. Thank you in advance!
Look at it this way:
$\displaystyle N(t) = 20 - \frac{30}{\sqrt{9 + t^2}}$

$\displaystyle N(t) = 20 - 30 \cdot (9 + t^2)^{-1/2}$

Now just apply the chain and power rules:
$\displaystyle N^{\prime}(t) = -30 \cdot -\frac{1}{2}(9 + t^2)^{-3/2} \cdot 2t$

$\displaystyle N^{\prime}(t) = 30t \cdot (9 + t^2)^{-3/2}$

$\displaystyle N^{\prime}(t) = \frac{30t}{(9 + t^2)^{3/2}}$

-Dan