Question

Can someone please help me with this equation.

Find the derivative of the function.

h(x) = 5/x^2 + 5/x^3

h'(x)= ??

Answer 1

Just rewrite h(x) using negative exponents :O should make it alil easier to see what's going on. Just the power rule from there.

\[\huge h(x)=5x^{-2}+5x^{-3}\]

Answer 2

-10x-15x^2?

Answer 3

Remember that when you SUBTRACT in the negative direction, then numbers should actually get larger. You're subtracting one from the power, here is what the first term will look like.

\[\huge (5x^{-2})'=5(-2)x^{-2-1}=-10x^{-3}\]

Answer 4

ok, then the second term would be -15x^-4

Answer 5

mhm c:

Answer 6

so -10x^-3-15x^-4?

Answer 7

yes, gj

Answer 8

@zepdrix can you help me with one more problem?

Answer 9

sure

Answer 10

\[ = (x^2-3x-1)^-7\]

Answer 11

So you'll take the derivative of the outermost function first, which is ( )^-7

Answer 12

Then we have to apply the chain rule :O multiply by the derivative of the inner function.

Answer 13

\[\large \frac{ d }{ dx }(x^2-3x-1)^{-7}=-7(x^2-3x-1)^{-8} [\frac{ d }{ dx }(x^2-3x-1)]\]

Answer 14

Confused? :O maybe the prime notation is a little easier to read, i dunno :3

Answer 15

just going over what you said.

Answer 16

ok i got it, so what would the next steo be?

Answer 17

\[\large \frac{ d }{ dx }(x^2-3x-1)\]
Hmm looks like we need to take this derivative still.

Answer 18

2x-3

Answer 19

Mmmmmmmmmmmmmm yah sounds good!

Answer 20

\[\huge -7(x^2-3x-1)^{-8}(2x-3)\]
and yah that should be our final answer, yay.

Answer 21

thanks a lot zep

Answer 22

one more? lol