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While attempting this problem with derivatives, I was bogged down in the notation and was unsure as to what was being derived. The problem goes as follows: Given $f(x)=x^3$, evaluate $\frac{d}{dx}f(x^3)$, at $x=2$ and state the general derivative. I tried solving for $f$ at $f(x^3)$ and then differentiating, but that yields $9x^8$, which doesnt seem right, considering $f'(x^3)=48x^2$, which should equal $192$. What am I doing wrong?

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  • $\begingroup$ I usually interpret $\frac{d}{dx}$ to behave like an operator (i.e. it can map functions to functions and brackets may be omitted) which makes sense as soon as you learn about weak derivatives. So $\frac{d}{dx}f(x^3)=\left(\frac{d}{dx} f\right) (x^3)$ for me. The interpretation given in the answer relies more on the limit-definition of the derivative which usually exclusively defines the derivative in a point $x$, so $\frac{d}{dx} f(x^3)=\frac{d}{dx}\left(f(x^3)\right)$ can be justified. This ambiguity is the reason I don't like this notation or would love to see people use more brackets... $\endgroup$
    – Piwi
    Oct 25, 2015 at 0:56

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Let $g(x)=x^3$, then $f(x^3)=f(g(x))$. Then use the chain rule $$(f(g(x)))' = f'(g(x))g'(x)$$

So we just need $f'(x)$ and $g'(x)$. $f'(x) = 3x^2$ and $g'(x) = 3x^2$ as well (because $f$ and $g$ happen to be the same function in this case). So then $$f'(g(x))g'(x) = f'(x^3)(3x^2) = (3(x^3)^2)(3x^2) = 9x^8$$

Or just figure out $f(x^3)$ explicitly: $h(x) = f(x^3) = (x^3)^3 = x^9$. Then $h'(x) = 9x^8$.

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