1. ## Power Rule Derivatives

I am having trouble with the power rule on derivatives. I understand that 1/x^5 is x^-5, but what about something like 16/x^2? The problem I'm working is square root x^3 - x^2 +16/x^2. With points (4, -7)

The way I am incorrrectly working it is 3x^1/2 - 2x + 16/2x^-1.

2. The derivative is a lineair operator, meaning that constants can be put up front and the derivative of a sum is the sum of the individual derivatives. So for 16/x², you just have:

(16/x²)' = 16*(1/x²)' = 16*(x^(-2))'

When there's a square root over the entire expression, use the chain rule. If it was only over the first term, write it as a power 1/2 and use a property of powers to put the exponents together: (a^b)^c = a^(bc).

3. ## thank you

Thanks for the help, that was giving me fits. I hope the rest of the problem was correct.

f'(x) = 3x^1/2 -2x + 16x^-2.

m=3(4)^1/2 -2(4) +16(4)^-2
m=12^1/2 - 8 + 64^-2 Now assuming this is right, do I do the problem like:

12(1/2) - 8 + 64(-2) or 1/12 - 8 + ??1/64^-1?

4. I'm a bit lost in your notation, was f(x) the following?

$\displaystyle f\left( x \right) = \sqrt {x^3 } - x^2 + \frac{{16}}{{x^2 }} = x^{\frac{3}{2}} - x^2 + 16x^{ - 2}$

Then find f'(x) by taking the derivative of each term.

5. You've gotten the problem right. I'm a bit confused on what to do with x^3/2, but I'll give it a shot.

f'(x) = 1/3x - 2x - 32x^-3? If you know of a site for me to do some extra studying on this, I would be more than willing to do some extra studying. I'm doing this course thru Independent Study and the book doesn't give many examples.

6. Originally Posted by becky
You've gotten the problem right. I'm a bit confused on what to do with x^3/2, but I'll give it a shot.

f'(x) = 1/3x - 2x - 32x^-3? If you know of a site for me to do some extra studying on this, I would be more than willing to do some extra studying. I'm doing this course thru Independent Study and the book doesn't give many examples.
Hello,

you nearly got it. But with:

$\displaystyle f\left( x \right) = x^{\frac{3}{2}} - x^2 + 16x^{ - 2}$ you'll get the first derivative:

$\displaystyle f' \left( x \right) = {3\over 2} \cdot x^{\frac{1}{2}} - 2x - 32x^{ - 3}$ = $\displaystyle {3\over 2} \cdot \sqrt{x} - 2x - \frac{32}{x^3}$

Now plug in x = 4 and you'll get:

$\displaystyle f'(4)={3\over 2} \cdot \sqrt{4} - 2\cdot 4 - \frac{32}{4^3}$ = $\displaystyle {3\over 2} \cdot 2 - 8 - \frac{1}{2}=-\frac{11}{2}$

Greetings

EB

7. Originally Posted by becky
...If you know of a site for me to do some extra studying on this, I would be more than willing to do some extra studying. I'm doing this course thru Independent Study and the book doesn't give many examples.
Hello,

try this one:
http://www.univie.ac.at/future.media/moe/

There exists an english version of this math online pages.

Especially the interactive tests are maybe of some interests for you.

Greetings

EB