Why does taking the logarithm of both sides not work?

[Hypercubes]

Homework Statement
Differentiate x=e^{-x}

The attempt at a solution
\ln{x}=\ln{e^{-x}}
\ln{x}=-x
\frac{d}{dx}\left(\ln{x}\right)=\frac{d}{dx}\left(-x\right)
\frac{1}{x}=-1

The correct answer is 1+e^{-x}. I know how to solve it that way; however, why is the above method wrong? I know there are usually two variables, but I can't see any mathematical errors in my method.

Thank you.

 

[tiny-tim]

Hi Hypercubes!

Differentiate x=e^{-x}
…
The correct answer is 1+e^{-x}.

Obviously a misprint, for "Differentiate x-e^{-x}"

 

[Hypercubes]

Well, the exact question was "explain why the equation e^{-x} = x has exactly one solution", and the answer goes on to define f(x) = x-e^{-x} and says "Since f'(x) = 1 + e^{-x}, and 1 + e^{-x} > 0 for all x, the function f is increasing..."

So they set it to 0 to get the derivative. However, why was the method I used not valid?

 

[D H]

Your expression, x=e^{-x} is just that -- an expression. You cannot treat it as a function. In particular, you can't differentiate both sides and expect something meaningful to pop out. What you can do is to rewrite it as x-e^{-x} = 0, and now the quantity on the right hand side f(x)=x-e^{-x} can be treated as a function. You aren't differentiating both sides of f(x)=0. Just one.

 

[Hypercubes]

Ah, I see. Thanks you both!

 

[Matterwave]

As a simple example. Take the equation 3x-2=x. This equation has 1 solution x=1. If you differentiate both sides though, you get 3=1 which is obviously not true.

This is because the undifferentiated equation is asking a fundamentally different question than the differentiated equation. The equation 3x-2=x is asking "When is the function f(x)=3x-2 equal to the function g(x)=x?" While the differentiated equation is asking "When is the slope of the function f(x) equal to the slope of the function g(x)?"