Question

how do i put this into wolfram? I can't solve them:(

Answer 1

you can type derivative <expression>

Answer 2

type

derivative ( 8^(4t) )/( t )

take note of the parenthesis

Answer 3

also be careful to notice that wolfram uses "log" as a natural log

Answer 4

wolfram uses log correctly :)

Answer 5

I'm used to log being base 10 by default and using "ln" for natural logs

Answer 6

so that caught me off guard at first

Answer 7

right , in advanced maths log is ln

Answer 8

it gave me this and its wrong:"( i have 2 more chances though

Answer 9

perhaps it wants "ln" instead of "log"

also see what happens when you distribute

Answer 10

Not sure. Sometimes computers can be really picky and wrong.

Answer 11

did you try logarithm differentiation

Answer 12

oh no :0 :"(

Answer 13

i tried it in my previous example and it was wrong... i have two of these problems and i get a 100% :( So i really want to do them:"(

Answer 14

well you can simplify that to get

\[\Large f'(t) = \frac{4096^t(12t\ln(2) - 1)}{t^2}\]

Answer 15

but that's not much of a simplification I guess

Answer 16

should i try that answer? after i try it it will give me a different example

Answer 17

sure lets see if that works

Answer 18

not sure what form they want it in

Answer 19

Yes, that was correct!!! :D

Answer 20

Can you help me with one more?

Answer 21

sure

Answer 22

so you'll need to type in

t^(3/2)*log( sqrt(t+5) )/log(2)

Answer 23

into wolfram?

Answer 24

correct

Answer 25

like this?:) thats my answer?

Answer 26

sorry I ask, i don't use it a lot and i don't wan't to learn to use it a lot, i like to learn math actually lol

Answer 27

and that simplifies to

\[f ' (t) = \frac {\sqrt {t} \left( 3\,\ln \left( t+5 \right) t+15\,\ln
\left( t+5 \right) +2\,t \right) }{ 4\ln \left( 2
\right)\left( t+5 \right) }\]

so yeah, definitely ugly...

Answer 28

Very ugly!! lol

Answer 29

Thank you :)

Answer 30

to do this by hand, you'd use the product rule
\[\large f(t) = t^{3/2}\log_{2}(\sqrt{t+5})\]
\[\large f'(t) = \frac{3}{2}t^{3/2-1}\log_{2}(\sqrt{t+5})+t^{3/2}\frac{1}{\ln(2)*\sqrt{t+5}}\]
\[\large f'(t) = \frac{3}{2}t^{1/2}\log_{2}(\sqrt{t+5})+t^{3/2}\frac{1}{\ln(2)*\sqrt{t+5}}*\frac{1}{2}(t+5)^{-1/2}\]
then you simplify that to get what wolfram is saying

Answer 31

Thank you for teaching me the steps, yes I actually tried it and got a mess and knew it was wrong right away lol

Answer 32

well it's definitely not a clean simple thing (unfortunately)

Answer 33

its easier to change the base first

Answer 34

also bring the exponent 1/2 down in front

log_2 (t + 5)^1/2 = 1/2 * log_2 ( t + 5) = 1/2 * ln ( t + 5) / ln 2

Answer 35

that is a much easier expression to take the derivative of

Answer 36

log_2 (t + 5)^1/2 = 1/ (2 * ln 2 ) * ln ( t + 5)

Answer 37

good point, that should help simplify things

Answer 38

Oh i see it, can i always change the base first? It won't change anything right?

Answer 39

no because

\[\Large \log_{b}(x) = \frac{\log(x)}{\log(b)}\]

\[\Large \log_{b}(x) = \frac{\ln(x)}{\ln(b)}\]

Answer 40

the logs on the right side can be any base

Answer 41

so that's why nothing changes

Answer 42

Oh!!!!! Okay okay :D i understand that!! Thank you, that makes so much sense

Answer 43

np

Answer 44

Good bye guys, I will now study for chemistry! Thank you :D

Answer 45

have fun

Answer 46

hold on look:(

Answer 47

it gave me this one now:"(

Answer 48

I got the same thing. Maybe they want you to use the * to mean multiply? hmm

Answer 49

ok now I get

\[\Large f'(t) = {\frac {\sqrt {t} \left( 3t\,\ln \left( t+4 \right)+12\,\ln
\left( t+4 \right) +2\,t \right) }{ 4\left( t+4 \right) \ln \left( 2
\right) }}\]

Answer 50

basically the same as before, just with different numbers

Answer 51

yea, but it always throws me off, sometimes they change one small thing and i make a huge mistake and get a whole different answer lol

Answer 52

let me try it:

Answer 53

well again you'll use the product rule and the chain rule will pop up when it comes to deriving the log

or you can simplify things first to make it a bit easier, then derive next

Answer 54

still wrong:(

Answer 55

yes its the same concept as the first one, correct?

Answer 56

try using * for multiplication symbols

Answer 57

yes, just the numbers have changed

Answer 58

oh wait, I missed a number, one sec

Answer 59

\[\Large f'(t) = \frac {\sqrt {t} \left( 3\,\ln \left( t+4 \right) t+12\,\ln
\left( t+4 \right) +2\,t \right) }{ 12\left( t+4 \right) \ln \left( 2
\right) }\]
ok got that now

Answer 60

I had the wrong log base before

Answer 61

it changed it to this

Answer 62

I can just change the #'s no?

Answer 63

third time is the charm (hopefully lol)

\[\Large f'(t) = \frac {\sqrt {t} \left( 3\,\ln \left( t+3 \right) t+9\,\ln
\left( t+3 \right) +2\,t \right) }{ 4\left( t+3 \right) \ln \left( 5
\right) }\]

is what I get

Answer 64

and yeah, the numbers are just changed

Answer 65

okay let me try my last chance lol

Answer 66

Perfect:) i got a 100% THANK YOU!

Answer 67

glad it finally accepted it lol

Answer 68

me too!!! Thanks to you!!