Finding derivative of function

RM5152

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Hello. I am confused by this problem question … I can’t figure out how to determine f from the information given. I will post my work this far… can anyone help please? ?40BCD677-415E-45B6-AD38-7FAB52593252.jpegABCEF467-9F52-42A6-913A-E86FCC2237F6.jpeg
 
Your work is incomplete!

You have that 1*f(0) = a. So solve for f(0).

You really need to use your knowledge of the 1 times table!

You have \(\displaystyle g'(x) = e^{-1}f'(x) + (-e^{-1})f(x).\) Now clean this up! Then use the fact that f'(x)=f(x)

A prerequisite for Calculus 1 is Algebra
 
You also need to learn some math grammar. To denote that you want to take the derivative of say e^(-x), you do not write e^(-x) (derivative). You write\(\displaystyle \dfrac {d}{dx}(e^{-x})\ or\ (e^{-x})'\)
 
If f'(x) = f(x) for all x what is g'(x)?

-Dan
T
Your work is incomplete!

You have that 1*f(0) = a. So solve for f(0).

You really need to use your knowledge of the 1 times table!

You have \(\displaystyle g'(x) = e^{-1}f'(x) + (-e^{-1})f(x).\) Now clean this up! Then use the fact that f'(x)=f(x)

A prerequisite for Calculus 1 is Algebra
Thanks for the help so far.. I have tried to find g’(x) but just can’t figure out how that could allow me to determine f as asked in question 6(b). this is my work this far.
 

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T

Thanks for the help so far.. I have tried to find g’(x) but just can’t figure out how that could allow me to determine f as asked in question 6(b). this is my work this far.
In your work, third line from the end, you write:

g'(x) = e^(-x) * (f(x) - f(x))

then you make a mistake. The next calculation should be :

g'(x) = 0

g(x) = K ..............................constant

You have to use algebra correctly to solve problems in Calculus.
 
T

Thanks for the help so far.. I have tried to find g’(x) but just can’t figure out how that could allow me to determine f as asked in question 6(b). this is my work this far.
If I asked you what 3 - 3 is, what would you tell me? So what is f(x) - f(x)?

-Dan
 
If I asked you what 3 - 3 is, what would you tell me? So what is f(x) - f(x)?
I see my mistake here. So g’(x) = e^-x (0) which equals 0 … so g’(x)=0. But I just don’t understand how this derivative can help me to find f. I know if g’(x) = 0 then g(x) = e^-x*f(x) must be a constant …. But I can’t figure out how to find f.
 
If I asked you what 3 - 3 is, what would you tell me? So what is f(x) - f(x)?

I see my mistake here. So g’(x) = e^-x (0) which equals 0 … so g’(x)=0. But I just don’t understand how this derivative can help me to find f. I know if g’(x) = 0 then g(x) = e^-x*f(x) must be a constant …. But I can’t figure out how to find f.
g(x) = e^-x*f(x) = K

use e^(-x) = 1/[e^x]

Algebra again!!!!
 
Did you read response #6 ?
Yes . I understand g(x) is K … a constant. So g(x) = e^-x*f(x) is a constant .. I still don’t know how to find f. I’m sorry I know I must obviously be missing something obvious but I just can’t get it
 
Yes . I understand g(x) is K … a constant. So g(x) = e^-x*f(x) is a constant .. I still don’t know how to find f. I’m sorry I know I must obviously be missing something obvious but I just can’t get it
Work with pencil/paper - don't just stare at the screen

e^-x*f(x) = K ............................now use e^(-x) = 1/(e^x) ................................. and algebra
 
Again, your work stops in the middle.
For example, you failed to complete e^(-x)(f(x)-f(x))=e^(-x)(0) =0.
Later on you had 1(a-a) = 1, which is wrong. Stop being lazy and write things down.
1(a-a) = 1(0) =0.
I truly believe that if you wrote 0 for a-a you would not have gotten the result of 1. Besides, you never had to do that calculation if you completed that g'(x) = e^(-x)(f(x)-f(x))=e^(-x)(0) =0.


Not simplifying equations is the number one mistake you are making. Just work on always asking yourself if you can simplify further.
 
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You are not finished, sorry.
You know that f(x) = Ke^x. Can you figure out what K equals.
 
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