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If $x=ye^y$, then explicit differentiation to find $\frac{dx}{dy}$, and the implicit differentiation to find $\frac{dy}{dx}$ yield consistent results.

Explain if the above line is True or False. (Ans: True)

But why? When we can clearly see using wolframalpha, explicit differentiation is $\dfrac{dx}{dy} = e^y \times (1 + y)$.

Implicit differentiation using wolframalpha, is $y'(x) = \dfrac{dy}{dx} = \dfrac{1}{e^y (y+1)}$

They seem clearly different, and not consistent result because they are reciprocal(?) of each other?

Mathew Mahindaratne
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nvs0000
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    The answers aren't meant to be equal, they're measuring different values. The result is consistent with $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}},$ which is a pretty well-known identity. – Stephen Donovan May 31 '21 at 03:59
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    $dy/dx$ and $dx/dy$ should be reciprocal each other. – Seewoo Lee May 31 '21 at 04:00
  • (1) So them being reciprocal is "consistent results"? /// (2) I thought consistent result means they are = each other. So it is my understanding of the question that is wrong..? – nvs0000 Jun 05 '21 at 11:44

1 Answers1

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If $f(x)$ and $g(x)$ are differentiable functions and $y = f(x) \cdot g(x)$, the derivative of the product $y$ is given by:

$$\frac{dy}{dx} = \frac{d}{dx}\left(f(x) \cdot g(x)\right) = f'(x)g(x) + f(x)g'(x)$$

Thus, for $f(y) = y$ and $g(y) = e^y$, then the differentiation of $x= ye^y$ gives you:

$$\frac{dx}{dy} = \frac{d}{dy}\left(ye^y\right) = e^y + ye^y = e^y(1 + y)$$

Thus, $\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}} = \dfrac{1}{e^y(1 + y)}$.

Doing implicit differentiation on $x= ye^y$ with respect to $x$:

$$\frac{dx}{dx} = \frac{d}{dx}\left(ye^y\right) = \frac{dy}{dx}e^y + y\frac{(de^y)}{dx} = \frac{dy}{dx}e^y + ye^y\frac{dy}{dx} = e^y(1 + y)\frac{dy}{dx}$$ $$\therefore \ e^y(1 + y)\frac{dy}{dx} = \frac{dx}{dx} = 1 \ \Rightarrow \ \dfrac{dy}{dx} = \dfrac{1}{e^y(1 + y)}$$

Mathew Mahindaratne
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  • So them being reciprocal is "consistent results"? I thought consistent result means they are = each other. So it is my understanding of the question that is wrong..? – nvs0000 Jun 05 '21 at 11:44
  • Please look at the comment by Stephen Donovan: "The answers aren't meant to be equal, they're measuring different values. The result is consistent with $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$, which is a pretty well-known identity." I think that says everything. – Mathew Mahindaratne Jun 05 '21 at 16:31
  • I see, I just got confused by the question is seem, thank you! Why is $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$ well known, where is it used..? Is there a reason why people won't just say $frac{dx}{dy}$ instead? – nvs0000 Jun 06 '21 at 02:52
  • It's from the chain rule. Read through this post and see if it helps. – Toby Mak Jun 06 '21 at 05:52
  • @Toby Mak: That is a good reference. +1 for you. – Mathew Mahindaratne Jun 06 '21 at 06:31