Derivative problem

[elquicko]

Find the derivative of: y = 1+ ln(t)
1 - ln(t)

I used the quotient rule:

1+ ln(t) * (1- ln(t))' - (1+ln(t))' * 1- ln(t)
(1-ln(t))²


>> 1+ln(t) * 1/t - (1/t * 1-ln(t))
(1-ln(t))²

 

[stapel]

Find the derivative of: y = (1+ ln(t))/(1 - ln(t))

I used the quotient rule:

y' = [1+ ln(t) * (1- ln(t))' - (1+ln(t))' * 1- ln(t)] / [(1-ln(t))^2]

>> y' = [1+ln(t) * 1/t - (1/t * 1-ln(t))] / [(1-ln(t))^2]

What is your question? Thank you!

 

[elquicko]

What is your question? Thank you!

Srry, just wondering if I did it correctly. I'm not confident in my skills right now.

 

[Ishuda]

You have it slightly backwards:
\(\displaystyle (\frac{f}{g})' = \frac{f' g - f g'}{g^2}\)
so we have
\(\displaystyle (\frac{1 + ln(t)}{1-ln(t)})' = \frac{[1+ln(t)]' [1-ln(t)] - [1+ln(t)] [1-ln(t)]'}{[1-ln(t)]^2}\)

=\(\displaystyle \frac{\frac{1}{t} [1-ln(t)] - [1+ln(t)] (-\frac{1}{t})}{[1-ln(t)]^2}\)
=\(\displaystyle \frac{\frac{1}{t} [1-ln(t)] + [1+ln(t)] (\frac{1}{t})}{[1-ln(t)]^2}\)
=\(\displaystyle \frac{\frac{2}{t}}{[1-ln(t)]^2}\)
=\(\displaystyle \frac{2}{t [1-ln(t)]^2}\)

You did, from what I can see,
\(\displaystyle (\frac{f}{g})' = \frac{f g' - f' g}{g^2}\)
which is the negative of the proper answer and also made a mistake in collecting terms/taking the derivative.

 

[ameri]

I think the answer should be:
2/t(ln(t)-1)^2

 

[lookagain]

I think the answer should be:
2/t(ln(t)-1)^2

ameri,
that is almost correct. You're missing grouping symbols around the denominator:

2/[t(ln(t) - 1)^2]


Or, if you stay in the style of the problem, the answer is also equal to:

2/[t(1 - ln(t))^2]



Note: (ln(t) - 1)^2 = (1 - ln(t))^2