# Simplifying this expression

https://www.quora.com/profile/Bravewarrior/p-152782085
I'm stuck on the very last line. I'm just confused on how they simplified the expression and got the final answer. Any help would be great!
Original post by pigeonwarrior
https://www.quora.com/profile/Bravewarrior/p-152782085
I'm stuck on the very last line. I'm just confused on how they simplified the expression and got the final answer. Any help would be great!
Presume youre ok when they sub for u,v and the derivatives. Then its just a case of
(a+b)/c = a/c + b/c
and noting that "a" is a fraction so a=d/e so
a/c = d/(ec)
as
(d/e)/(c/1) = (d/e)*(1/c) = d/(ec)
and cancelling the common (x+1)
Original post by mqb2766
Presume youre ok when they sub for u,v and the derivatives. Then its just a case of
(a+b)/c = a/c + b/c
and noting that "a" is a fraction so a=d/e so
a/c = d/(ec)
as
(d/e)/(c/1) = (d/e)*(1/c) = d/(ec)
and cancelling the common (x+1)
Thank youuu I finally get it now! 🙂
Original post by pigeonwarrior
Thank youuu I finally get it now! 🙂
Just as an aside, youd avoid the "complicated" simplifying at the end if you noted that youre differentiating
ln(x)(x+1)^(-1)
using the product rule so
1/(x(x+1)) - ln(x)/(x+1)^2
Sometimes the quotient rule (which is a particular case of the product rule) makes things more complex, though it obviously gives an equivalent expression. The usual quotient rule expression for the derivative of f/g is equvalent to
f'/g - fg'/g^2
which is what they wanted here.
(edited 1 month ago)
Original post by mqb2766
Just as an aside, youd avoid the "complicated" simplifying at the end if you noted that youre differentiating
ln(x)(x+1)^(-1)
using the product rule so
1/(x(x+1)) - ln(x)/(x+1)^2
Sometimes the quotient rule (which is a particular case of the product rule) makes things more complex, though it obviously gives an equivalent expression. The usual quotient rule expression for the derivative of f/g is equvalent to
f'/g - fg'/g^2
which is what they wanted here.
That is such a good point! It would make it much easier for me, thank you so much 🙂