Domain of a derivative

  • Thread starter mbrmbrg
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  • #1
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Can the domain of f'(x) be larger than the domain of f(x)?

Here's my work, as per the rules:smile:

The problem I'm working on states
f(x) = x + √x
and says to find f'(x), the domain of f(x), and the domain of f'(x).

I found that f'(x)=1, so I want to say that its domain is all real numbers. But the domain of f(x) is only (0, ∞).

Hence my question in the first line.
 

Answers and Replies

  • #2
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Look at your differentiation again. You only calculated the derivative of f(x) = x.
 
  • #3
You need to differentiate Sqrt(x) too, as mentionned by Hammie and then see for what values of x D[f(x) = x + Sqrt(x)] is defined.

By that I mean check for what values of x the denominator of D[f(x)] does not equal zero. If the denominator of D[f(x)] equals zero, then D[f(x)] is not defined for these particular values of x because dividing by zero is one of the few "forbidden" operations in math.

To answer your question, the domain of D[f(x)] can include the domain of f(x). For example f(x) = ln(x)
The domain is ]0, +infinity[ and D[f(x)] = 1/x had R/{0} as domain.

Hope it helps
 
  • #4
HallsofIvy
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[tex]\frac{df}{dx}_{x=a}[/tex]
is defined as
[tex]lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}[/tex]
If f(a) is not defined, then f '(a) is not defined either.
 
  • #5
486
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I re-did the problem and came up with f'(x) = 1 + [1/(2√ x)].
Is that the right answer? (It's an even problem, so the book doesn't have the answer...)
Thank you!
 
  • #6
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Looks good to me. So, what does that say about the domain?
 
Last edited:
  • #7
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That means that my original question is irrelevant for this problem, b/c the domain of f'(x) is x=0 or x>0, which the domain for f'(x).

Thanks everyone, both for checking my work and answering my original question!
 
  • #8
HallsofIvy
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No, it's not.
[tex]f(x)= x+ \sqrt{x}[/tex]
is defined for x= 0 but
[tex]f '(x)= 1- \frac{1}{2\sqrt{x}}[/tex]
is not!

And the answer to your original question is "No, the domain of f ' cannot be larger than the domain of f, but it can be a proper subset."
 
  • #9
Thank you a lot HallsofIvy for correcting my mistakes. I had mistaken the derivative ( the linear application wich associates the function to it's linear approx. ) and the function it represents.
 

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