# Thread: Another Derivative

1. ## Another Derivative

Hello All,

Here is the problem:

$\displaystyle y=\sqrt[n]{\frac{1}{x^m}}$

Here is what I got:
$\displaystyle \frac{dy}{dx}=\frac{-mx}{n}^\frac{-m-n}{n}$

Here is what the answer I was given:
$\displaystyle \frac{dy}{dx}=\frac{-mx}{n}^\frac{-m+n}{n}$

I don't understand how the exponent could be -m+n??? When you take the derivative of a power, you are supposed to subtract one, correct? If so, doesn't
$\displaystyle \frac{-m}{n}$ -1 = $\displaystyle \frac{-m-n}{n}$

2. Yes, your answer is the correct one.

3. Originally Posted by HallsofIvy
Yes, your answer is the correct one.
Thats what I thought. I asked just in case I was missing something or overlooking an algebra error.

Thanks.

-db

4. Hello, dbakeg00!

Differentiate: .$\displaystyle y\;=\;\sqrt[n]{\frac{1}{x^m}} \;=\;x^{-\frac{m}{n}}$

Here is what I got: .$\displaystyle \frac{dy}{dx}\;=\;-\tfrac{m}{n}x^{-\frac{m}{n}-1} \;=\;-\tfrac{m}{n}x^{\frac{-m-n}{n}}$ . . . . Right!

Here is what I was given: .$\displaystyle \frac{dy}{dx}\;=\;-\tfrac{m}{n}x^{\frac{-m{\color{red}+}n}{n}}$ . . . . Wrong!

"They" must have a typo.

Your work is correct!