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I'm sorry for the uncertainty. Thanks to your advices, now I can point out what I wonder.

$$When \quad e^x=y$$ 1. $${\partial F\over\partial y}=e^{-x}{\partial F\over\partial x}$$ 2.$${\partial^2 F\over\partial y^2} =(e^{-2x}{\partial^2 \over\partial x^2} \mathbf{-e^{-2x}{\partial \over\partial x}})F$$

$$where \quad F(y,z) \quad is \quad a \quad function \quad of \quad y \quad and \quad z, \quad and \quad is \quad a \quad function \quad that \quad is \quad of \quad class \quad C^{\infty}$$ $$ and \quad where \quad variable \quad z \quad is \quad independent \quad of \quad x \quad and \quad y.$$

In 1, I understood it analogically with the following:

$$de^{x}={\partial e^{x} \over \partial x}dx=e^{x}dx$$

But 2 is very confusing. I can't derive the result above step by step. Though the_architect thankfully let me know that product rule is used in 2, I still don't know how to do it manually.

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    @DonAntonio I tried to get it better though it might be funny yet. I want to know how the two results are derived – 2017-02-01
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    Most certainly the calculation done in the picture is $\partial_y^2 = (e^{-x} \partial_x)(e^{-x} \partial_x) = e^{-x} ( e^{-x} - e^{-x} \partial_x ) \partial_x$ that means using the product equation for differentiation. – 2017-02-01
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    @Philippe8 wHat do you have to begin with? The equation (function) $\;y=e^x\;$ ? and then you need to evaluate the derivative of the *inverse function* of this function? – 2017-02-01
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    $\frac{\partial}{\partial x}$ shouldn't be written like that; you should always write $\frac{\partial f(x,y)}{\partial x}$ with an unambiguous function $f(x,y)$ with here $f(x,y)=0 \iff y=e^x$ wich could be something like $f(x,y)=1-ye^{-x}$ (but it doesn't frankly fulfill your constraints) or $f(x,y)=1-y^2e^{-2x}$ or... – 2017-02-01
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    You should learn to type in Latex (https://en.wikibooks.org/wiki/LaTeX/Mathematics) – 2017-02-01
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    @DonAntonio I edited it. Now I wish my intention is apparent. – 2017-02-02
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    @Philippe8 There are already three intents of answer. After you editer your question I don't have anymore any idea what you want. First of all, what in the world is that $\;F\;$ in point (1) ?? Where does it come from? Before that you only have what seems to be a function $\;y\;$ of the one variable $\;x\;$. Of course, it could be that we are in three dimensional space so you'll get a surface in $\;\Bbb R^3\;$ defined by $\;e^x=y\iff F(x,y):=e^x-y=0\;$ ...but there's no explanation at all. This is a very poorly worded question... – 2017-02-02
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    @DonAntonio Oh still there was something missed. Sorry for bothering you. I edited it. And the thing I want to know is step-by-step derivation of 1 and 2, especially 2. Thank you for replying several times though the question seemed like nonsense. – 2017-02-02

2 Answers 2

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For example:

$$y=e^x\iff x=\log y\implies \left(\log y\right)'=\frac1{\left(e^{\log y}\right)'}=\frac1{e^{\log y}}=\frac1y$$

If you wanted something else then I couldn't guess it correctly...And I can't guess what partial derivatives have to do here.

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If you are treating $y$ as a function of $x$ (i.e. $y=y(x)$) I don't see why you try to do something like this. Partial derivatives are used for functions with multiple variables: $$f_1(x,y)=e^x,\quad f_2(x,y)=y,\quad f_3(x,y)=e^x-y$$ for example.

Maybe your question comes from abuse of notations often found in Mechanics e.g. $$f(x,y) \text{ where } y=e^x$$ so $${df \over dx}={\partial f \over \partial x}{dx \over dy}+{\partial f \over \partial y}.$$ If this is the case, DonAntonio's answer would help you