Question

If a general function is...

Answer 1

Let \[f^{n}(a)\] denote the nth derivative of f at a.

If \[f(x) = \ln(2x^{2} +x -1)-\ln(x+1), \]

then \[f^{98}(\frac{1}{2}+\frac{\sqrt{2}}{2}) =? \]

Answer 2

I did \[\ln \frac{ 2x^{2}+x-1 }{ x+1 }\]

Answer 3

the numerator is factorable

Answer 4

you'll just get the form ln (ax+b)

Answer 5

and you know its n'th derivative ?

Answer 6

to take derivative I use product rule and chain rule. \[\ln \frac{ (x-1)(2x-1) }{ x-1 }\]

Answer 7

why not simply cancel out x-1 ?

Answer 8

This becomes \[\ln(2x-1)\]

derivative = \[\frac{1}{2x-1}*\frac{d}{dx}(2x-1)\]

Answer 9

yes, do you know the n'th derivative formula or have been taught ?

Answer 10

I guess you keep taking the derivative until you see a pattern? or is there another way @hartnn

Answer 11

yes, same thing.

Answer 12

see i you get this pattern
\(\huge \dfrac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n}\)
for ln |ax+b|

Answer 13

hmm. how could I see that.

Answer 14

1st derivative = 2/ (2x-1)
2nd derivative = - 2^2 / (2x-1)^2

so , there will be alternate + and - signs, (-1)^n-1 will account for that

every time 2 will get multiplied, 2^n will account or that
denominator is quite obvious

Answer 15

i suggest you take that formula for ln |ax+b| as standard and memorize it...there won't be enough time to derive it everytime...

Answer 16

That's right..I need to practice find these patterns,,,,

Answer 17

so the general formula in my case is \[\frac{ -1^{n-1}(n-1)! 2^{n} }{ (2x-1)^{n} }\]

Answer 18

yes, just plug in n=98
x = 1/2 +1/sqrt 2

Answer 19

2x-1 is just sqrt 2

Answer 20

-97! * 2^49
is what i get, did u get same ?