Question

g(t)=(t-4)ln(t+5)
find g'(t) & g"(t)

Answer 1

i thought i understood this problem but no...
i thought ln(t+5) turns to (1/(t+5)) and (t-4) turns to 1 so the answer would be 1/(t+5) for g'(t) but its not...

Answer 2

@ash2326

Answer 3

sorry had to tag you lol

Answer 4

we have
\[g(t)=t-4 \ln(t-5)\]
using product rule
we get
\[g'(t)=(t-4)\frac{1}{t-5}+ \ln (t-5)\]

Answer 5

You try for g''(t)

Answer 6

would it be 1/(t+5) + 1/(t+5) so 2/(t+5)??

Answer 7

No, for the first term use division rule
(u/v)'= (vu'-uv')/v^2
\[t-4/t-5= ( (t-5)(1)-(t-4)(1))/(t-5)^2

Answer 8

im not seeing it. can u fill in the steps u skipped.

Answer 9

sorry if im being complicated

Answer 10

We have
\[g'(t)=\frac{t-4}{t-5}+\ln (t-5)\]
let's find g''(t)
\[g''(t)=\frac{(t-5)\frac{d}{dt} (t-4)-(t-4)\frac{d}{dt}(t-5)}{(t-5)^2}+\frac{1}{t-5}\]
we get
\[g''(t)=\frac{(t-5)\times 1-(t-4) \times 1}{(t-5)^2}+\frac{1}{t-5}\]
\[g''(t)=\frac{t-5-t+4}{(t-5)^2}+\frac{1}{t-5}\]
we get
\[g''(t)=\frac{-1}{(t-5)^2}+\frac{1}{t-5}\]

Answer 11

i thought it was PLUS 5
(t+5)

Answer 12

@mariomintchev did you understand?

Answer 13

why is it t-5 not t+5?

Answer 14

Oh I didn't see that it's t+5 . Sorry
replace (t-5) by (t+5) everywhere

Answer 15

so 9/(t+5)^2 + 1/(t+5) is the final answer

Answer 16

yeah :D

Answer 17

my hw website isnt accepting that answer

Answer 18

It can also be written as
\[g''(t)= \frac{9}{(t+5)^2}+\frac{1}{t+5}=\frac{9}{(t+5)^2}+\frac{t+5}{(t+5)^2}\]
or
\[g''(t)= \frac{t+14}{(t+5)^2}\]
try this

Answer 19

nope

Answer 20

:(

Answer 21

any other ways i could put it in?

Answer 22

@ash2326
you online?

Answer 23

\[g(t)=(t-4)ln(t+5)\]
\[g(x)=t\ln(t+5)-4\ln(t+5)\] to start, then
\[g'(t)=\ln(t+5)+\frac{t}{t+5}-\frac{4}{t+5}\] or if you prefer
\[g'(x)=\ln(t+5)+\frac{t-4}{t+5}\] n replete
\[g''(x)=\frac{1}{t+5}+\frac{9}{(t+5)^2}\]
or if you want to add the fractions
\[\frac{t+4+9}{(t+5)^2}\]

Answer 24

actually this is exactly what ash wrote above

Answer 25

@amistre64
this is the final one that we've been struggling with

Answer 26

any ideas?

Answer 27

Sorry mario. Im not good with calc. :(

Answer 28

:(

Answer 29

didja get it yet?

Answer 30

nooooooooooooo

Answer 31

lol

Answer 32

everyones getting similar answers but my hw site isnt accepting anything i put in. ive put in over 20 answers. ha

Answer 33

this one aint asking for values

Answer 34

i can redo whats already been done; but that wont really change things will it

Answer 35

unless you gave the wrong problem, theres not much more to do

Answer 36

g(t)=(t+4)ln(t+5)

Answer 37

g'(t) is ln(t+5)+(t+4)/(t+5)

Answer 38

i need g''(t)

Answer 39

then do it again :) nothings stopping you from applying derivative again to it

Answer 40

show me what you get and lets see where you having the issues at

Answer 41

everywhere

Answer 42

been on this problem forever

Answer 43

yes; everywhere

Answer 44

Ok I got: \[g'(t)=\ln{(t+5)}+\frac{t-4}{t+5}\]

Answer 45

\[g'(t) = ln(t+5)+\frac{(t+4)}{(t+5)}\]
derive it thru again for g''

Answer 46

lets use the last edition of the problem this time tho ...

Answer 47

and \[g''(t)=\frac{t+14}{(t+5)^2}\]

Answer 48

the t-4 is from the first post

Answer 49

is not t-4?

Answer 50

i end up with 9/(t+5)^2 + 1/(t+5)

Answer 51

what is the function you want to derivate?

Answer 52

its PLUS 4. SORRY!

Answer 53

well in that case I got \[g''(t)=\frac{t+2}{(t+5)^2}\]

Answer 54

and \[g'(t)=\ln{(t+5)+\frac{t+4}{t+5}\]

Answer 55

lets see what I get :)

\[g'(t) = ln(t+5)+\frac{(t+4)}{(t+5)}\]
\[g''(t) = ln(t+5)'+[(t+4)(t+5)^{-1}]'\]
\[g''(t) = (t+5)^{-1}+(t+4)'(t+5)^{-1}+(t+4)(t+5)' ^{-1}\]
\[g''(t) = (t+5)^{-1}+t(t+5)^{-1}-(t+4)(t+5)^{-2}\]
\[g''(t) = ((t+5)+t(t+5)-(t+4))(t+5)^{-2}\]

\[g''(t) = \frac{(t+5)+t(t+5)-(t+4)}{(t+5)^{2}}\]
\[g''(t) = \frac{t+5+tt+5t-t-4}{(t+5)^{2}}\]
\[g''(t) = \frac{t^2 +5t+1}{(t+5)^{2}}\]

Answer 56

\[(t+4)'\neq t\]\@amistre64

Answer 57

yeah, was just getting to that :)

Answer 58

t+6 up top

Answer 59

wonderful piece of work ruint by a lousy t :)

Answer 60

yeah a lot of latex work =/

Answer 61

but what is the answer you get then?

Answer 62

I think it'd be
\[g''(t)=\frac{1}{t+5}+\frac{1}{(t+5)^2}\]
we get
\[g''(t)=\frac{(t+6)}{(t+5)^2}\]

Answer 63

10-4 = 6 soo
\[g''=\frac{t+6}{(t+5)^2}\]

Answer 64

i love you all!!!

Answer 65

thats the answer

Answer 66

Yeah, you're awesome!