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The question is: Let $F(x) = f^2(g(x))$. If $g(1)= 2, g'(1)= 3, f(2) = 4$, and $f'(2) = 5$, find $F'(1)$.

$$ F'(x) = f'(f(g(x)) * (f \circ g)'(x) = f'(f(g(x)) * f'(g(x)) * g'(x)$$

put $1$ into $F'(x)$:

$$F'(1) = f'(f(g(1)) * f'(g(1)) * g'(1) = f'(f(2)) * f'(2) * 3 = f'(5) * 5 * 3$$

I can only solve the problem into this step. Is there any mistake I make? I cannot find a way to simplify the part of $f'(f(g(x))$. Thank you!

Sebastiano
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1 Answers1

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You lost a factor $2$ $$ F'(x)=(f^2(g(x)))'=2f(g(x))(f(g(x))'=2f(g(x))f'(g(x))g'(x). $$

Marius S.L.
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  • Both your and OP's formulas might be correct depending on the interpretation of the ambiguous "$f^2$" as either the square power or the square composition. I do feel like your interpretation is the one that was expected by whoever asked the question though considering the fixed values given in the question. – Bruno B Dec 10 '22 at 11:54
  • so f^2(x) mean (f(x))^2 instead of f(f(x)), right? – davislui Dec 10 '22 at 11:57
  • Depends on the context, which is why it's ambiguous here. Most of the time I've seen it's usually reserved for $f \circ f$ only, but sometimes (often with $\cos$, $\sin$ and co) it designates $x \mapsto f(x)^2$, like I'd assume is the case here. (When both significations can be involved I like to write $f^{\circ 2}$ but I don't know if it's standard in any way) – Bruno B Dec 10 '22 at 11:59
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    That is my reading. The power to the 2 at functions means squaring the function value in 9 out of 10 cases. I would have expected the notation $F(x)=f(f(g(x)))$ otherwise. In the end, it depends on the context. If we were in algebras of functions, then it wouldn't be clear at all. I assumed a context that matches the difficulty of the problem. In ordinary calculus, notations like $\sin^2, , ,\log^2$ etc. mean a square. Complexity theory would become a total mess if people wrote $\log^2$ for $\log \log.$ – Marius S.L. Dec 10 '22 at 12:02
  • Got it. Thanks for your clear explanation – davislui Dec 11 '22 at 06:34