Question

Find the derivative.

Answer 1

\[\frac{ t }{ (1-t)^3 }\] @zepdrix

Answer 2

Make sure you're able to get the initial setup correctly :D Cause it all goes downhill from there if you mess that part up hehe.
\[\huge \frac{ (t)'(1-t^3)-(t)(1-t^3)' }{ (1-t^3)^2 }\]

Answer 3

i thought it's the quantity cubed, not just the t?

Answer 4

isn't the derivative of t just 1?

Answer 5

Was the whole quantity supposed to be cubed? Hmm it doesn't look like it based on the way you have it written :D

Yes t gives you 1.

Answer 6

Yes the whole quantity is cubed, that's what i wrote in the original problem i think o_O

Answer 7

Oh you did write that :D man im off my game tonight!

Answer 8

\[\huge \frac{ (t)'\left[(1-t)^3\right]-(t)\left[(1-t)^3\right]' }{ \left[(1-t)^3\right]^2 }\]

Answer 9

Ok ok ok ok does that look better? my bad :3

Answer 10

Giving youuuuu something to this effect :O
\[\huge \frac{ (1)\left[(1-t)^3\right]-(t)\left[-3(1-t)^2\right] }{ \left[(1-t)^3\right]^2 }\]

Answer 11

The primes are to show where we still NEED to take a derivative :D Hopefully the notation isn't confusing.

See how in this next step, I applied the derivative to those 2 terms, and dropped the primes?

Answer 12

i was missing the power of 2 of the derivative of (1-t)^3 gah. all right i'll check again. And yea i deleted the comment b/c i'm just so stubborn lol sorry

Answer 13

haha XD

Answer 14

Wait how am i supposed to simplify this then? do i use pascal's triangle and do it the long way?

Answer 15

Ummmm I suppose you could do some factoring like this, which gives you a nice cancellation. See how I pulled the square thing out of both terms in the top?
\[\huge \frac{ (1-t)^2\left[(1-t)+3t\right] }{ (1-t)^6}\]

Answer 16

no haha i'm actually very confused o_O

Answer 17

Hmm so the top simplifies to this, right?
And then we need to figure out how to proceed.
\[\huge \frac{ (1-t)^3+3t(1-t)^2 }{ (1-t)^6 }\]

Answer 18

Maybe this will help :) Maybe not, but maybe.. so let's try it.
Let (1-t)=x
\[\huge \frac{x^3+3t\cdot x^2}{x^6}=\; \frac{x\cdot x^2+3t\cdot x^2}{x^6}\]

Answer 19

See how they both have an x^2 in them? :o

Answer 20

\[\huge =\frac{x^2(x+3t)}{x^6}\]

Answer 21

Is the factoring part a little confusing? :D

Answer 22

the first one isn't it -3t?

Answer 23

\[\huge (1-t)^3\]The derivative of this gave us:\[\large 3(1-t)^2(1-t)'=\quad -3(1-t)^2\]

Answer 24

And that term was being subtracted, so it changed the minus to a positive, right? :o

Answer 25

Make sure you're doing the correct one first :D
You always differentiate the TOP of the fraction first,

You said the "first" term, so I wasn't sure if maybe you made a mistake on that :D

Answer 26

what i'm staring at right now is this: \[(1-t)^3-3t(1-t)^2 \] for the numerator.

Answer 27

I think you forgot to apply the chain rule when you took the derivative of the second term.
A -1 should pop out when you differentiate the most inner function (1-t).
Changing it to:\[(1-t)^3+3t(1-t)^2\]

Answer 28

oh ok. then how do i go from there?

Answer 29

\[(1-t)^3+3t(1-t)^2=\quad (1-t)\cdot (1-t)^2+3t(1-t)^2\]
Factoring out a (1-t)^2 from each term gives you:\[=(1-t)^2\left((1-t)+3t\right)\] on top.

Answer 30

ohhh i see. wow that solved the entire problem...you, sir, are a genius. thank you so much haha. i'll be back to annoy you however with more problems :P

Answer 31

hah :D I'm prolly gonna be busy for a bit XD gotta study for a test tomorrow :D just do the @ thing and I'll try to take a look if i have a minute :3

Answer 32

yup definitely, thanks! i'll try not to disturb you :P