Question

dy=(x^4-1)/(3x) Find deriavative??

Answer 1

dy or y?

Answer 2

use quotient rule
\[(\frac{f}{g})' =\frac{f'g -fg'}{g^{2}}\]

Answer 3

\[y=\frac{1}{3}x^3+\frac{1}{3}x^{-1}\]

Answer 4

sorry its not dy its y

Answer 5

\[\frac{d(ax^n)}{dx}=nax^{n-1}\]is the only thing you need to use

Answer 6

use it

Answer 7

what is the derivative of \[\frac{1}{3}x^3\]?

Answer 8

@fakharabbas786 we are not here to do it for you, we are here in hopes that we can show you how to do it.

Answer 9

do you know the derivative of x^3?

Answer 10

3x^2

Answer 11

so what is the derivative of (1/3)x^3

Answer 12

its the same thing multiplied by (1/3)

Answer 13

what is it then?

Answer 14

its confusing me

Answer 15

so when we take the derivative of something like (1/3)x^3 we take the derivative of x^3 and as you said this is 3x^2 but we still have the (1/3) out front. so its (1/3)*3*x^2 = x^2

Answer 16

\[3\frac{1}{3}x^2\]

Answer 17

so what is the derivative of \[\frac{1}{3}x^{-1}\]?

Answer 18

same thing as before, we bring down the -1 and then de-increment the exponent by 1
\[(-1)\frac{1}{3}x^{-2}=\frac{-1}{3}x^{-2} = \frac{-1}{3x^2}\]

Answer 19

zzrock3r u r confusing me

Answer 20

\[so\space the\space derivative\space of\\\frac{x^4-1}{3x}=\frac{1}{3}x^3+\frac{1}{3}x^{-1}\\is\\x^2-\frac{1}{3x^2}\]

Answer 21

first we take the derivative of (1/3)x^3 and get x^2
then we take the derivative of (1/3)x^(-1) and get (1/(3x^2))

Answer 22

then we add them together

Answer 23

tell me what part is confusing

Answer 24

you two are lame.....