Question

Derive:

30. f(s) = 5^s * e^s

Very confused here, wouldn't e^2 be a constant? shouldn't that be 0?
then it would just be ln(5) * 5^s? where or why am I wrong?

Answer 1

is it e^2 or e^s ?

Answer 2

e^s

Answer 3

Hint :-
\(\large (f(x)g(x))'=f(x)'g(x)+f(x)g'(x)\)

Answer 4

\[\large 5^s = e^{s\ln 5}\]

Answer 5

we haven't learned that yet, we can't use the product or quotient rule

Answer 6

ok then use what ganesh said :)

Answer 7

Can you describe what you did please

Answer 8

I have no idea why you combined them or the reasoning behind the answer

Answer 9

cos differentiating a single exponent function is easier than differentiating the product :

\(\large (e^{ax})' = a e^{ax}\)

Answer 10

I see that you didn't get why \(\large 5^s\) equals \(\large e^{s\ln 5}\)

Answer 11

exactly, I don't know why that's equivalent, I get that 5^s*e^s = (5e)^s then it would be the ln(5e) * 5e^s

Answer 12

that looks perfect !!

Answer 13

is that correct? then what would the next step be? that's where I'm stuck, can I factor it ?

Answer 14

you have used \(\large (a^x)' = a^x \ln a\) formula right ?

Answer 15

yes

Answer 16

\[\large (5^s * e^s)' = \left((5e)^s \right)' = (5e)^s \ln (5e)\]

there is no next step, you're done !

Answer 17

really? but (BOB) back of the book says something else

Answer 18

what does it say

Answer 19

it has (1+ln(5)) 5^s e^s?

Answer 20

haha the book simplified \( \ln(5e)\)

Answer 21

\(\ln (ab) = \ln a + \ln b\)

Answer 22

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa!!!!!!!!!!!! jesus christ, I'm going to have to borrow your brain for my test

Answer 23

lol, in test we can leave the answer as \((5e)^s \ln (5e)\)
you won't loose marks for not simpilifies this

Answer 24

ln(5e) looks simple to me than (1 + ln5)

one term is better than two terms :P

Answer 25

BOB and I don't get along very well sometimes. thank you very much. =).

Answer 26

yw :) good luck wid the test !

Answer 27

thanks =)