# Derivatives problem

• Jul 15th 2007, 05:51 PM
starswept
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!
• Jul 15th 2007, 05:56 PM
topsquark
Quote:

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:
$N(t) = 20 - \frac{30}{\sqrt{9 + t^2}}$

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

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

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

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

-Dan