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Mathematics 18 Online
OpenStudy (anonymous):

dy/dt of y=(ln(2t))/(5t)

OpenStudy (anonymous):

y = a/b ; y' = (ba' - ab')/b^2

OpenStudy (anonymous):

Yes I know it's the quotient rule but somewhere I am messing up... can't find my error

OpenStudy (anonymous):

I get ((ln(2t)*5)-((1/2t)*2)(5t))/(5t^2) which is incorrect, sorry for all the parenthesis it's how I have to enter it on on the website haha

OpenStudy (anonymous):

trying denominator*(derivative of numerator) - (numerator)*derivative of denominator

OpenStudy (anonymous):

Yeah I understand that I'm just messing up somewhere I suppose... Can't find it though it's really frustrating!

OpenStudy (anonymous):

deriv of ln(2t) = (1/2t)*2 right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Idk then.. my answer looks 100% right to me but it isn't! :(

OpenStudy (anonymous):

i got (1-ln2t)/5t^2. is that right?

OpenStudy (anonymous):

Nope sadly

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle y=\frac{\ln(2t) }{5t} }\) \(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{\frac{1}{t}(5t)+5\ln(2t) }{(5t)^2} }\) \(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{5+5\ln(2t) }{25t^2} }\) \(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{1+\ln(2t) }{5t^2} }\)

OpenStudy (xapproachesinfinity):

firstly the way you started is incorrect you must abide the order when you do the quotien rule unlike the product rule where you can start with any

OpenStudy (solomonzelman):

I differnetiated the top function, except I wrote the + instead of -

OpenStudy (solomonzelman):

\(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{1-\ln(2t) }{5t^2} }\)

OpenStudy (anonymous):

thats what i got

OpenStudy (solomonzelman):

F(x)/G(x) = { F`(x) G(x) - F(x)G`(x) } / G^2(x)

OpenStudy (xapproachesinfinity):

@ineptAtMath yes your answer is correct

OpenStudy (anonymous):

Okay you were right ineptatMath. I needed to add a parenthesis

OpenStudy (solomonzelman):

and why is derivative of any ln(Ct) with respect to t is 1/t \(\large\color{black}{ \displaystyle (~\ln(Cx)~)'=\frac{1}{Cx}\times C }\) or \(\large\color{black}{ \displaystyle (~\ln(Cx)~)'=(~\ln(x)+\ln(C)~)'=\frac{1}{x}+0=\frac{1}{x} }\)

OpenStudy (solomonzelman):

So in ln(Cx) (had a typo in the beginning) is just a vertical shift of ln(x) , C units up

OpenStudy (anonymous):

(1-ln2t) needed to be (1-ln(2t)) lol

OpenStudy (anonymous):

Thanks for the explanation and help guys! Going to go over what you've sent so I can see what I did wrong

OpenStudy (solomonzelman):

;)

OpenStudy (anonymous):

Oh sorry about that!

OpenStudy (solomonzelman):

abt what /

OpenStudy (solomonzelman):

?

OpenStudy (anonymous):

the lack of parenthesis in what i wrote was confusing

OpenStudy (solomonzelman):

oh, if you write a space, it is most fine, but yes, don't leave your parenthesis out, becuase not everyone will see the typo behind it.

OpenStudy (solomonzelman):

I hand-written my essay once, and my "second" looked like "secont" (in 8th grade) And that brought my paper from 98 to 93.

OpenStudy (anonymous):

duly noted! ^^ And omg..... just one misspelled word dropped you down 5%? Harsh

OpenStudy (solomonzelman):

no, for that mistake in 8th grade, it is just like not knowing what integral is in calculus 2.

OpenStudy (solomonzelman):

it is not as low level mistake to assume a typo, but not as high to go over...

OpenStudy (solomonzelman):

anyway.... \(\Huge\bf\rlap{\color{red}{G\ o\!}}{\color{blue}{\; {G\ o\ }}}\)\(\Huge\bf\rlap{\color{red}{o\ d\ }}{\color{blue}{\; {o\ d\ }}}\) \(\Huge\bf\rlap{\color{red}{L\ u\!}}{\color{blue}{\; {L\ u\ }}}\)\(\Huge\bf\rlap{\color{red}{c\ k\ }}{\color{blue}{\; {c\ k\ }}}\) \(\Huge\bf\rlap{\color{red}{~!}}{\color{blue}{\; ~{!}}}\)

OpenStudy (anonymous):

Haha okay! Good luck to you too :)

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