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OpenStudy (anonymous):

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

OpenStudy (cwrw238):

yes

OpenStudy (cwrw238):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohhhhhhhhhhhhhhhhhh got it!

OpenStudy (anonymous):

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

OpenStudy (cwrw238):

(lnt)^2 (1) is correct ( i made a mistake there)) yes its ( ln t)^2 not ln^2 t

OpenStudy (anonymous):

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

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the answer is 2lnt+lnt^2

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thats where im stuck

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

agreed?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (cwrw238):

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'

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

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

OpenStudy (cwrw238):

* derivative of 3x is 3

OpenStudy (anonymous):

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

OpenStudy (cwrw238):

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

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