# Domain of a derivative

Can the domain of f'(x) be larger than the domain of f(x)?

Here's my work, as per the rules 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.

Look at your differentiation again. You only calculated the derivative of f(x) = x.

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

HallsofIvy
Homework Helper
$$\frac{df}{dx}_{x=a}$$
is defined as
$$lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$
If f(a) is not defined, then f '(a) is not defined either.

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!

Looks good to me. So, what does that say about the domain?

Last edited:
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!

HallsofIvy
$$f(x)= x+ \sqrt{x}$$
$$f '(x)= 1- \frac{1}{2\sqrt{x}}$$