Question

\[f(t)= \frac{ 2 }{ \sqrt{t} }+\frac{ 6 }{ \sqrt[3]{t} }\]

Answer 1

noted

Answer 2

What are you trying to do with that expression?

Answer 3

take the derivative

Answer 4

let's try and rewrite this equation otherwise we will be using quotient rules all over the place and that would be nasty

Answer 5

So it will be 2(t^-1/2) + 6(t^-1/3)

Answer 6

yeah and we can take product rule.. must easier.. remember constants don't have derivatives.. they go to 0 for example f(x) = 1 ----> f'(x) = 0

Answer 7

so product rule means that we leave the first term alone ... deal with the second + leave the second term alone deal with the first
fg'+f'g

Answer 8

so let's try it ^-^
2(t^-1/2) so who do we leave alone first in this one?

Answer 9

the (t^-1/2)?

Answer 10

umm sort of.. if I was doing the product rule backwards... but let's stick to the rules.. it's easier

Answer 11

so let's let f= 2 and g = t^(-1/2)
if the product rule is fg'+f'g who is term that we leave alone for fg'

Answer 12

2

Answer 13

yes. so we take the derivative of g which is take the derivative of t^(-1/2)

Answer 14

it would be -1/2t^-3/2

Answer 15

yes!
so now we have 2 (-1/2t^-3/2) which is ...

Answer 16

we still need to simplify XD !

Answer 17

just multiply 2 with the -1/2