Question

find d4y/dx4,if y=log(2x+5)

Answer 1

we need to find the 4th derivative

Answer 2

have you found the first ?

Answer 3

And I guess you log(2x+5) is understood to be log_(10)(2x+5)

Answer 4

the first will be 2(2x+5)^-1

Answer 5

is that it?

Answer 6

Is the base of the log e?

Answer 7

if so yes

Answer 8

do i have to be finding the derivative one after the other?

Answer 9

yep
(Also is the base of that log e?)

Answer 10

or is that log natural?

Answer 11

is there no formula to find all the four at once?

Answer 12

sometimes you are asked to derive a formula
we can derive one here by observing pattern

Answer 13

natural.

Answer 14

Ok so you have f'=2(2x+5)^-1
now we need f''=

Answer 15

f"=-1x2x2(2x+5)^-2

Answer 16

is it?

Answer 17

what is all those x's?

Answer 18

oh are some of them multiplication signs and the only variable x you have is in that parenthesis next to the 2?

Answer 19

multiplication

Answer 20

You actually can see a pattern here and skip finding the third derivative

Answer 21

notice:
\[f'=2(2x+5)^{-1} ; f''=-1 \cdot 2^2 (2x+5)^{-2}\]

Answer 22

let's take a guess at the fourth derivative from this

Answer 23

you know the the derivatives are sign alternating

Answer 24

I'm going to rewrite the first derivative one more time so it is more obvious
\[f'=2^1(2x+5)^{-1} ; f''=-1 \cdot 2^2 (2x+5)^{-2} \]

Answer 25

do you see a pattern yet?

Answer 26

wait let me rewrite one more time to make more obvious
\[f^{(1)}=2^1(2x+5)^{-1} ; f^{(2)}=-1 \cdot 2^2 (2x+5)^{-2}\]

Answer 27

f'"=-1.-2.2^3(2x+5)^-3

Answer 28

the 3rd derivative should be negative
and shouldn't have 2*2^3 but just 2^3

Answer 29

shouldn't be negative*

Answer 30

oh wait is that the 4th derivative so many '

Answer 31

that still wouldn't be the 4th derivative

Answer 32

no, 3rd

Answer 33

so you weren't able to see the pattern from the first two derivatives
\[f^{(1)}=(-1)^{-1+1} \cdot 2^{1}(2x+5)^{-1}\]
\[f^{(2)}=(-1)^{-1+2}\cdot 2^2 (2x+5)^{-2}\]
\[f^{(3)}=(-1)^{-1+3} \cdot 2^3(2x+5)^{-3}\]
now do you see the pattern?

Answer 34

the (-1) to some power was thrown in there to get the alternating part

Answer 35

notice the things that change and the things that don't change

Answer 36

omg i made an error

Answer 37

forgot to bring down the 2 I think

Answer 38

\[f^{(3)}=(-1)^{-1+3} \cdot 2^{4}(2x+5)^{-3}\]

Answer 39

so you were right
i'm sorry

Answer 40

okay

Answer 41

have got the pattern

Answer 42

thank you

Answer 43

you do
so you could find the nth derivative? :)

Answer 44

spoiler below
____for nth derivative:
\[f^{(n)}=(-1)^{-1+n} \cdot 2^n \cdot (n-1)! \cdot (2x+5)^{-n}\]

Answer 45

There is your formula if you really wanted to just find one for this problem.

Answer 46

Also honestly there are too many unnecessary formulas to memorize like this one

Answer 47

I say it is unnecessary because this can easily be derived by other formulas

Answer 48

you can't remember them all
our brain is limited

Answer 49

yes

Answer 50

You did very well @dzie
so you get a medal too

Answer 51

thank you myininaya