Question

Derive 4t^(-1/3)

Answer 1

what do you want to derive?
do you mean differentiate?

Answer 2

Use the power rule

If \(\Large y = x^n\) then \(\Large y' = n*x^{n-1}\) where n is some constant

Answer 3

You also use the coefficient rule. So you pull out the coefficient of 4 and ignore it temporarily while you derive t^(-1/3). Once you are done deriving, don't forget to put it back in.

Answer 4

derive is to differentiate?

Answer 5

yes in terms of calculus

Answer 6

I have never heard derive to mean differentiate

Answer 7

derive I thought meant to like derive the quadratic formula by completing the square
i thought i meant to basically to find

Answer 8

true, derive in a general sense means to take something already established or given and create a new rule, theorem, etc

Answer 9

but with calculus it also means to differentiate

Answer 10

According to this one site I found Germans use derive to mean differentiate

Answer 11

I got -1.3t^(-1/3) @jim_thompson5910 but its not right

Answer 12

perhaps because you are using the approximate form

Answer 13

it is true that 4 times -1/3 = -4/3 = -1.3 approximately, but I would stick with -4/3 as the final coefficient

Answer 14

oh and you forgot to subtract the exponent

Answer 15

notice how it's n-1 in the exponent (not just n)

Answer 16

oops that's what I meant but it didn't work

Answer 17

n = -1/3

n-1 = ???

Answer 18

im doing my COW assignment

Answer 19

idk what COW stands for

Answer 20

but if n = -1/3, then what is n-1 equal to?

Answer 21

calculus on the web

Answer 22

ah gotcha

Answer 23

are you able to calculate n-1 ?

Answer 24

-4/3

Answer 25

so the answer is \[\Large -\frac{4}{3}t^{-4/3}\]

Answer 26

it's a coincidence that the coefficient and exponent match up like this

Answer 27

ohh thanks

Answer 28

np