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Suppose we have a composite function: $f(g(h(x)))$, and we want $\frac{\partial f}{\partial h}$.

By the chain rule $\frac{\partial f}{\partial h} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h}$.

My question is: what about any occurrences of $x$? Do we have to express all the $x$ as functions of $h$? Or even write the $x$s in terms of $g$? Does it make a difference?

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For example $f = g\times (x^2+1)$, and $g = h^2; h=1/x^2$. Then

1) $\frac{\partial f}{\partial h} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial h} = (x^2+1)\times 2h$,

or write $f = h^2\times(x^2+1)$, and

2) $\frac{\partial f}{\partial h} = 2h\times(x^2+1)$,

or write $f = h^2\times(1/h+1) = h + h^2$, and

3) $\frac{\partial f}{\partial h} = 1+2h$.

(1) and (2) give the same result, and $x$ is not acted on. What about case (3)?

user243205
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    I've asked this many times in comments, never got an answer. How do people define derivative with respect to a function? – Git Gud May 25 '15 at 00:06
  • $f = f(g(h(x))) = g(h(x))*(x^2 + 1)$. This is just an example, I invented some forms for $g$ and $h$. – user243205 May 25 '15 at 00:16
  • @GitGud I think that is the heart of the issue here. Are they meant to be functional derivatives in the chain rule? – user243205 May 26 '15 at 01:32
  • My opinion is that the symbol $\frac{\partial f}{\partial h}$ has no meaning and I have no idea what is actually meant. – Git Gud May 26 '15 at 08:55

1 Answers1

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$$f = g(h(x))*(x^2 + 1)$$ $$h(x)=\dfrac{1}{x^2}$$ $$f = g(h(x))*\dfrac{(h(x)) + 1}{(h(x))}$$ Let $h(x)=t$, $$f = g(t)*\dfrac{t + 1}{t}$$ $$\dfrac{\partial f}{\partial t}=\dfrac{\partial g}{\partial t}\dfrac{t + 1}{t}-g(t)*\dfrac{1}{t^2}$$

vidhan
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  • Your f is now wrong, it should read $f = g(h(x))*\frac{h(x)+1}{h(x)}$. Iḿ more interested in the general principle rather than this particular example. Your answer is that we must first write the occurrences of $x$ in terms of h(x), and then start differentiating as per the chain rule? – user243205 May 25 '15 at 03:33