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If $h(x) = f(g(x))$, then what does $f'(g(x))$ mean? $dh/dg$ or $dh/dx$?

If $h(x)$ is a composite function such that $h(x)= f(g(x))$, then what does the derivative of $f(g(x))$ with respect to $g(x)$ mean?

ryang
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Anandhu_S
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3 Answers3

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as I can see you are confused with the notation in the chain rule: well $f' (g(x))$ is equal to say : compute $g(x)$; let's say $g(x) = k$ now compute $f'(k)$ note that : $[f(g(x))]' = h'(x) = g'(x)f'(g(x)) \neq f'(g(x)) $

Michael Hoppe
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ilia varnasseri
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Regarding the title-"If h(x)=f(g(x)), then what does f′(g(x)) mean? dh/dg or dh/dx?" f' is actually a bit vague. It is not 100% clear. Technically, $h'( g(x) ) \text{ should be} = dh/ dg$ But since 90% of the time, f' means df/ dx, the possibility the original author made a mistake is quite high or a tad sloppy, especially if the person writing it is a student in Calculus I.

Whenever it's unclear which variable we are taking the derivative with respect to g or x, the notation dh/dg or dh/ dx is preferred to make it abundantly clear.

An alternate notation to make it clear is $D_g [f( g(x) )]$ or $D_x [f( g(x) )]$ Bottom line- avoid ambiguity in math notation whenever possible.

Regarding the question, "If h(x) is a composite function such that h(x)=f(g(x)), then what does derivative of f(g(x)) w.r.t g(x) mean?" the most common context is the chain rule. $$ \frac{df}{dx} = \frac{df}{dg} *\frac{dg}{dx}$$ $\frac{df}{dg}$ means take the derivative of f, with respect to g, $\frac{df}{dg}$. In practice, I teach this with the phrase,
"A. Take the derivative of f, or outer function,
B. recopy the inner with no changes" Then chain rule finishes by multiplying $\frac{dg}{dx}$

nickalh
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  • Arrgghh, my comment is a response to the original title, currently "If h(x)=f(g(x)), then what does f′(g(x)) mean? dh/dg or dh/dx?" Not the slightly different question underneath. – nickalh Oct 03 '21 at 07:02
  • I think the edit changes the question substantially. Typically, students learning chain rule are simultaneously learning the significance of the denominator in df/dg or in df/dx. – nickalh Oct 03 '21 at 07:22
  • Let g(x) be $x^2$ , and f(x) be sin(x), therefore
    $h(x) =f(g(x))=sin(x^2)$. And by chain rule,
    $h'(x)=dh/dg . dg/dx$ ; which would be equal to $2x.cos(x^2)$ ..My question is : why would $h'(g(x))$ be equal to $dh/dg$.? Can you explain its meaning using the functions I said above?     – Anandhu_S Oct 03 '21 at 11:49
  • @nickalh I'm sorry about the mistaken Question edit, where I'd mistaken and appended Rahul's Answer as the OP's own. Have reverted the edit. The orginal Question title though remains moved to the Question body. – ryang Oct 03 '21 at 18:05
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let f be some continuous function. Then f'(x) denotes the derivative of f wrt x i.e df/dx. In the similar way, we have f'(g(x) means df/dg. It's just a matter of notation. We can calculate those derivative via chain rule. If you are beginner of calculus and have some doubts like these, then videos by khan academy on calculus are really helpful. https://youtube.com/playlist?list=PL19E79A0638C8D449

RAHUL
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