Question

find the derivative, y=t(lnt)^2, natural log=ln* do i use the product rule?

Answer 1

yes

Answer 2

dy/dt = t* d(ln t(^2 / dt + ln t * dt/dt

Answer 3

ok wa am i doing wrong this is wa im getting, (t)(2lnt/lnt^2)+(lnt^2)(1)...

Answer 4

as far as setting up the problem with the product rule i mean

Answer 5

ohhhhhhhhhhhhhhhhhh got it!

Answer 6

i was putting lnt^2 instead of (t)(lnt^2) righ?

Answer 7

(lnt)^2 (1) is correct ( i made a mistake there))

yes its ( ln t)^2 not ln^2 t

Answer 8

so (t)(2lnt/t(lnt^2))+(lnt^2)(1) is correct?

Answer 9

d [(lnt)^2] / dt = 2 ln t * 1 / t = 2 ln t / t

Answer 10

dy/dt = t * 2 ln t / t + (ln t) ^2 * 1
= 2 ln t * (ln t)^2

Answer 11

sorry its
2 ln t + (ln t)^2

= ln t ( 2 + ln t)

Answer 12

wait wouldnt it b t*2lnt/t(lnt^2)? cuz of f'x/fx for logs?

Answer 13

the answer is 2lnt+lnt^2

Answer 14

hmm - i think thr answer is 2 ln t + (ln t)^2

Answer 15

not sure what u mean by f'(x) / fx for logs

Answer 16

how did you get 2lnt/t instead of 2lnt/t(lnt^2)?

Answer 17

thats where im stuck

Answer 18

the trickiest part is differentiating (ln t)^2 where we use the chain rule

Answer 19

lets do it this way:
let y = (ln t)^2
let u = ln t
then y = u^2 and and u = ln t
so dy/du = 2u and du/dt = 1 /t
so dy/dt = dy/du * du/dt = 2u * 1/t
= 2 ln t / t
and t * 2 ln t / t = 2 ln t

Answer 20

agreed?

Answer 21

yes

Answer 22

question, when do i use the chain rule, like is there a "standard" problem to use the chain rule?

Answer 23

so 2 ln t is first part of the derivative

the second part is (ln t)^2* dt/dt
= (ln t)^2 * 1
= (ln t)^2

finally derivative is

2 ln t + (ln t)^2

note (ln t)^2 = ' the log of t all squared'
whereas ln t^2 = ' the log of t squared'

Answer 24

right u use chain rule when you have a 'function within a function'

for example sin 3x :- 3x is a function of x and sin 3x is a function of 3x
sqrt(2x - 3) :- 2x - 3 is a f of x and sqrt(2x - 3) = function of 2x-3

Answer 25

so differentiating sin 3x:

derivative of 3x is and derivative of 'outer' function is cos 3x
so derivative of original is
3 * cos 3x = 3 cos3x

Answer 26

* derivative of 3x is 3

Answer 27

ok, i think i got it, i just gotta practice it. thank u!!!!!!!!!!

Answer 28

yes - it needs practice - but first try putting u = inner function as i did but after a while you can do these without resorting to that