Chapter 9, exercise 10 Michael Spivak's Calculus

Question

The problem is this. "Find $f'(x)$ If $f(x)=g(t+x)$ and $f(t) = g(t+x)$. The answers will not be the same."

The function $g(t+x)$ is a constant one, otherwise $f(x)$ would not be a function. Hence $f(x)$ is constant as well. Therefore $f'(x)$ is zero. But the solution to this exercise goes as follows:

"$f'(x)= g'(t+x)$ by problem 8(a), $f'(t)= g'(t+x)$ again by problem 8(a). Hence $f'(x)= g'(2x).$"

Just a note. The problem 8(a) asks if $f(x)=g(x+a)$ then $f'(x)=g'(x+a)$. Which is true just by definition of the derivative.

My problem is of course why are not the same and why is $f'(x)=g'(2x)$?. Guess since this is constant another solucion could have been $f'(x)=g'(\pi x)$. But I am quite insecure about this. Hope you can help me.

José Carlos Santos · Answer 1 · 2017-06-13T18:59:47.937

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Since $f'(t)=g'(t+x)$, then $f'(x)=g'(x+x)=g'(2x)$. No, $f'(x)=g'(\pi x)$ would not have been another solution.

edited Jun 13 '17 at 18:59

answered Jun 13 '17 at 18:57

José Carlos Santos

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@JoséCarlosSantos wouldn't f'(x) be 2g'(2x)? – mutantacule Apr 11 '21 at 10:18
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@kosii No, it will not. – José Carlos Santos Apr 11 '21 at 10:25
@JoséCarlosSantos I see now why, thank you. – mutantacule Apr 18 '21 at 08:46
@JoséCarlosSantos Can you explain why not? If $f(t) = g(t+x)$ then can't we say that $f(x) = g(2x)$, from which we obtain $f'(x) = 2g'(2x)$. Wherein lies the fallacy? – EE18 Aug 17 '22 at 17:08
@EE18 When we write that $f(t)=g(t+x)$, what that means is that $t$ is a variable and $x$ is a constant. Yes, $f(x)=g(2x)$, but that's true for the constant $x$. It doesn't meant that, for every $x$, $f(x)=g(2x)$. If it meant that, then, yes, you could deduce that $f'(x)=2g'(2x)$. – José Carlos Santos Aug 17 '22 at 21:03
Ah, I see. Thank you for that explanation! – EE18 Aug 20 '22 at 17:16

Chapter 9, exercise 10 Michael Spivak's Calculus

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