Two young mathematicians discuss derivatives as functions.

Check out this dialogue between two calculus students (based on a true story):

Devyn
Riley, I might be a calculus genius.
Riley
Yeah? Explain this one to me.
Devyn
Let me first ask you a question. Say you have a function, like \(f(x) = x^2\), and you want to know \(f'(3)\). Do you plug in the number \(3\) before or after you find the derivative?
Riley
Hmmmm. Well, my next step is usually
\[ f'(3) = \lim _{h\to 0}\frac {f(3+h)-f(3)}{h}. \]
So I guess before.
Devyn
Aha! I think you’re wasting time. You see I write
\[ f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(x)}{h}. \]
and it means that I can look at the derivative of my function at any point. So, I plug in the \(3\) after I’ve found the derivative.
Riley
That does seem like a pretty genius move. But doesn’t working with \(x\), instead of numbers, make all of this more difficult?
Devyn
Not at all. Let’s do the problems both ways, at the same time:
Riley
Whoa. So now the derivative is a function. Wait, what’s its domain? Its range?
Suppose you have a function \(f\). Which of the following are true?
The domain of \(f'\) is equal to the domain of \(f\). The range of \(f'\) is equal to the range of \(f\). The domain of \(f'\) is a subset of the domain of \(f\). The range of \(f'\) is a subset of the range of \(f\).
Find \(g'(2)\) for \(g(x) = x^2 + 1\) using both methods described above.
\[ g'(2) = \answer {4} \]
2025-01-06 20:00:25