Question

Derivatives

Answer 1

\[2te ^{t} - \frac{ 1 }{ \sqrt{t} }\]

Answer 2

How many?

Answer 3

how many what?

Answer 4

How many derivatives?

Answer 5

one

Answer 6

Never mind, lame attempt to be funny . . .
Ok, so it's just some power rule with a little product rule.

Are you familiar with those differentiation rules?

Answer 7

The \(2t \cdot e^t\) is a pretty easy product to use product rule on ("first times the derivative of the second, plus the second times the derivative of the first.")

Answer 8

oh isn't 2 a coefficient though? and lol i gave you the best response for your attempt

Answer 9

(Thanks!)

Yes, you can treat the 2 as a constant coefficient, it just goes along for the ride.

Answer 10

and the derivative of e^t is just e^t correct?

Answer 11

Always.

Answer 12

alright, so i got 2t * e^t + 2* e^t + t^1/2

Answer 13

Very close.
Still need to differentiate the \(-t^{-1/2}\)

Answer 14

(-t^1/2)-1

t^-1/2

-1/squareroot of t?

Answer 15

\[\large \frac{d}{dx} -t^{-1/2} = \frac{1}{2}t^{-3/2}\]

Answer 16

Sorry, should be \(\large \frac{d}{dt}\) at the beginning there.

Answer 17

well i didn't distribute the negative sign. so i just saw it as (t^1/2)^-1 originally

Answer 18

ohhh ok.

Answer 19

\[1/2^{3} \sqrt{t}^{2}\]

Answer 20

It's still a -3/2 exponent on the t, so it's in the denominator.

Answer 21

yea, i put it back in fraction form

Answer 22

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

Answer 23

Ok, but make sure in final form it is \[\large \frac{1}{2\sqrt{t^3}}\]

Answer 24

or t, not x

Answer 25

exponent -3/2 is reciprocal of square-root of 3rd power.

Answer 26

that's weird lol, i've been doing that, and i get it right....