MATH HELP NOW

[Multani]

so just bring it up, make its power in negative and apply product rule

yes, but then you have to use chain rule too

 

[Jf Thunder]

yes, but then you have to use chain rule too

yes that too

 

[PoKeMon]

View attachment 24085
plzzz help, so i can post more questions

Ahh my loving subject...use to solve these 10 yrs back.

 

[Jf Thunder]

Ahh my loving subject...use to solve these 10 yrs back.

do it again

 

[BDforever]

do it again

i did these type of math at college and i was good in math, let me recall.

 

[Jf Thunder]

i did these type of math at college and i was good in math, let me recall.

recall please

 

[BDforever]

recall please

need little help, show me similar type of question and answer lol

 

[Jf Thunder]

need little help, show me similar type of question and answer lol

answer, i dont have an answer

 

[BDforever]

answer, i dont have an answer

ok tell me is it differential or integral math

 

[Jf Thunder]

ok tell me is it differential or integral math

i think its differentiation

 

[BDforever]

i think its differentiation

i need a calculas book to recall anyway give me tonight, i will show it to my friends and solve it
edit: i have found my calculas book, give me some time, i will solve it

 

[Ra'ad]

I think part (a) is differentiation,
and part (b) is to prove its turning point (stationary point).

So differentiation has been done in the previous posts, resulting in
y’ = ((1/x)(x+1) - (1)(lnx)) / (x+1)^2

In part (b), we find the stationary point. Since at the stationary point, gradient is zero, so putting our derivative equal to zero results in the form: lnx = (x+1)/x (swap lnx and x, and u get the form)

Next you prove that the point is between x=3 and x=4.
Stationary point gradient = 0
The gradient before and after the stationary point have opposite signs.
Try putting in values: f'(3) and f'(4), they should have different signs. And thats it.