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I was revising chain rule and I made up a problem to write down in my notes that uses it at least two times. Here it is, if a function $\zeta(x) = (z(x))^2$ where $z(x) = x + f(x), f(x) = \ln(g(x))$ and $g(x) = \frac{1}{2}x^2$ then $\zeta'$ or $\frac{d\zeta}{dx}$ is defined as, \begin{align*} \zeta'(x) & = \frac{d\zeta}{dz}\times \frac{dz}{df}\times \frac{df}{dg} \times \frac{dg}{dx}\\ \zeta'(x) & = 2(z(x))z'(x) \\ & = 2(z(x))(1 + f'(x)) \\ & = 2(z(x))(1 + (\ln(g(x)))') \\ & = 2(z(x))\Big(1 + \Big(\frac{1}{g(x)}\Big)g'(x)\Big) \\ & = 2(z(x))\Big(1 + \Big(\frac{1}{g(x)}\Big)x\Big) \\ \end{align*}

Did I get it right?

scribe
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    Be confident in your work. And yes, it is correct. – cxx May 24 '20 at 06:20
  • Correct from the 2nd line to the end. You did not use the 1st line, & it's good that you didn't, else you would be stuck at the term $\frac {dz}{df}=\frac {d(x+f(x))}{df(x)}=??$ – DanielWainfleet May 24 '20 at 09:33
  • @DanielWainfleet read the comments under Matheus's answer. That is exactly what got me. Could you please explain perhaps in an answer how can one correctly write the first line? – scribe May 24 '20 at 16:24

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Have a little more confidence! Your work is completely correct.

About writing it in Leibniz notation: we have $\zeta(x) = \varphi(y)$, where $\varphi$ is defined by $x \mapsto x^2$ and $h$ by $y \mapsto h(y) = y + f(y)$. Therefore: $$ \begin{aligned} \frac{\mathrm{d}\zeta}{\mathrm{d}x} &= \frac{\mathrm{d} \varphi}{\mathrm{d}y} \frac{\mathrm{d}y}{\mathrm{d}x} \\ &= \frac{\mathrm{d} \varphi}{\mathrm{d}y} \left(1 + \frac{\mathrm{d}f}{\mathrm{d}y} \right) \end{aligned}$$

And from here on you can use the definitions of $f$ and $g$ to expand the last equality in terms of $f$ and $g$, but I'd rather leave it at just the first one.

Matheus Andrade
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  • Could you help me write it in Leibniz form? I know it'll be something like $\frac{d\zeta}{?} \frac{df}{dg} \frac{dg}{dx}$. But what will go in place of the question mark? – scribe May 24 '20 at 06:30
  • $\operatorname{d}f$? So it cancels out while multiplication – Alexey Burdin May 24 '20 at 06:40
  • @scribe I've added an edit that I think will help – Matheus Andrade May 24 '20 at 06:45
  • Okay I see, I edited my post. Though I have managed to confuse myself a little about $\frac{dz}{df}$. $\frac{dz}{df}$ says that we are taking the derivative of $z$ with respect to $f$. Since it is with respect to $f$, should we not treat the $x$ in $z(x) = x +f(x)$ as constant, yielding $\frac{dz}{df} = f'(x)$? I know this is wrong and my instinct the first time was right. But now I can't pin point to why $\frac{dz}{df} = f'(x)$ is incorrect. – scribe May 24 '20 at 07:04
  • @scribe It's incorrect because $z$ is not to be treated as a function of $f$, rather of $x$. It doesn't depend on $f$, because $f$ is a fixed function, unlike $x$, which takes values on the whole domain of $z$. – Matheus Andrade May 24 '20 at 07:13
  • I suppose then I am confused about how could we write a function in the denominator of Leibniz notation because if I understand correctly in $df / dx$, $f$ is always a function and $x$ a variable in it (hence differentiating $f$ with respect to $x$). Could you have a look at the first line of my edited equations and see if it is still correct as some pointed out that it is not. – scribe May 24 '20 at 16:29