Question

using the product rule, find the derivative: (x^-5+3)(x^-3+5)

Answer 1

i keep coming up with 3x^-9 as the first part of the answer but the correct answer is supposed to contain -8x^-9 and I don't see how to arrive at this

Answer 2

\[(-5x^{-6})(x^{-3}+5)+(-3^{-4})(x^{-5}+3) \rightarrow -5x^{-9}-25x^{-6}-3x^{-9}-9x^{-4} \]
\[\rightarrow -8x^{-9}-25x^{-6}-9x^{-4}\]

Answer 3

ya so i have all that except for some reason i multiplied exponents and didnt notice

thanks a bunch

Answer 4

You're welcome.

Answer 5

can i throw another at you?

Answer 6

x^2 + 8x + 3 all over / sqrt X

I don't get doing square roots with this stuff

Answer 7

A square root is just a fractional exponent, an inverse exponent is just a negative exponent. So you can rewrite all of that (if its easier) as \[\frac{x^{2}+8x+3}{\sqrt{x}} \rightarrow \frac{x^{2}+8x+3}{x^{\frac{1}{2}}} \rightarrow (x^{2}+8x+3)(x^{-\frac{1}{2}})\]

Answer 8

ya so how do i multiply x^2 by x^-1/2, this is the part i dont get

Answer 9

to me thats just -x

Answer 10

When you multiply numbers with exponents you simply add them together, so you would have 2-1/2 or 3/2

Answer 11

i knew you were going to say that

Answer 12

:)

Answer 13

so then the derivative of x^-1/2 is ????

Answer 14

\[x^{-\frac{1}{2}}\rightarrow -\frac{1}{2}x^{-\frac{3}{2}}\rightarrow -\frac{1}{2x^{\frac{3}{2}}}\rightarrow -\frac{1}{2\sqrt{x^{3}}}\]

Answer 15

thats ridiculous

Answer 16

so is my answer 3x^2+8x-3 all over 2x^3/2

Answer 17

Most likely, I didn't do the math so I will trust that you did. :)

Answer 18

i obviously suck at math

Answer 19

thanks for your help, much appreciated

Answer 20

Not a problem. :)