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$$ f(x) =\dfrac{1}{1 + e^{-x}} $$

The derivative of the above function is:

$$ {f}'(x) =\dfrac{e^{-x}}{(1 + e^{-x})^2} $$

But when I try to use the definition of the derivative:

$$ h = \Delta x $$

$$ {f}'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} $$

My solution turned out this way:

$$ =\lim_{h \to 0} \dfrac{\dfrac{1}{1 + e^{-(x+h)}} - \dfrac{1}{1 + e^{-x}}}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{1 + e^{-x}-(1 + e^{-(x+h)})}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) }}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{1 + e^{-x}-1 - e^{-(x+h)})}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) }}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{e^{-x}- e^{-x-h}}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) }}{h} $$

If $h \to 0$:

$$ =\lim_{h \to 0} \dfrac{0}{(1 + e^{-x})^2 } = 0 $$

Where am I making a mistake to get to the result

$$ {f}'(x) =\dfrac{e^{-x}}{(1 + e^{-x})^2} $$

Could you help me?

rafaelcb21
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  • Note $\lim\limits_{h \to 0} \dfrac{{e^{-x}- e^{-x-h}}}{h} = e^{-x}$ – Henry Feb 08 '23 at 23:41
  • Ok, my mistake putting $ h^2 $, but how does this $ e^{-x}- e^{-x-h} $ become $ e^{-x} $, could you show me? – rafaelcb21 Feb 08 '23 at 23:48
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    $e^{-x}-e^{-x-h} = e^{-x}(1-e^{-h})$. Do you know the limit of $(1-e^{-h})/h$ as $h$ goes to zero? – eyeballfrog Feb 08 '23 at 23:52
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    $e^{-x}- e^{-x-h}$ does not become $e^{-x}$ as $h \to 0$. But$ \dfrac{{e^{-x}- e^{-x-h}}}{h}$ does; it is the derivative of $-e^{-x}$ – Henry Feb 08 '23 at 23:54

1 Answers1

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The answer step by step, the way was to use L'Hopital's rule when there is an indeterminacy in the limit like $$ \frac{0}{0} or \frac{\infty}{\infty} $$

step by step:

$$ f(x) =\dfrac{1}{1 + e^{-x}} $$

$$ h = \Delta x $$

$$ {f}'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{1}{1 + e^{-(x+h)}} - \dfrac{1}{1 + e^{-x}}}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{1 + e^{-x}-(1 + e^{-(x+h)})}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) }}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{1 + e^{-x}-1 - e^{-(x+h)})}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) }}{h} $$

$$ =\lim_{h \to 0} \dfrac{\dfrac{e^{-x}- e^{-x-h}}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) }}{h} $$

$$ =\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{(1 + e^{-(x+h)})\cdot (1 + e^{-x}) } \cdot\dfrac{1}{h} $$

$$ =\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{h} \cdot\dfrac{1}{(1 + e^{-(x+h)})\cdot (1 + e^{-x})} $$

Here we have an indeterminacy:

$$ =\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{h} = \dfrac{e^{-x}- e^{-x-0}}{0} = \dfrac{0}{0} $$

So we can use L'Hopital's rule differentiating each function:

$$ \dfrac{d}{dh} \left(e^{-x} - e^{-x-h}\right) = \dfrac{d}{dh} (e^{-x}) - \dfrac{d}{dh} (e^{-x-h}) = 0 + e^{-x-h} = e^{-x-h} $$

$$ \dfrac{d}{dh} (h) = h = 1 $$

$$ \dfrac{\dfrac{d}{dh} \left(e^{-x} - e^{-x-h}\right)}{\dfrac{d}{dh} (h)} = \dfrac{e^{-x-h}}{1} $$

Substitute by zero

$$ =\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{h} = \dfrac{e^{-x-0}}{1} $$

$$ =\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{h} = e^{-x} $$

We do the distributive on the denominator:

$$ =\lim_{h \to 0}\dfrac{1}{(1 + e^{-(x+h)})\cdot (1 + e^{-x})} $$

$$ \lim_{h \to 0}\dfrac{1}{1 + e^{-x} + e^{-(x+h)} + e^{-(x+h)}\cdot e^{-x}} $$

We simplify:

$$ \lim_{h \to 0} \dfrac{1}{1 + e^{-x} + e^{-x-h} + e^{-2x-h}} $$

we can substitute for 0 that will not occur indetermination:

$$ \lim_{h \to 0} \dfrac{1}{1 + e^{-x} + e^{-x-0} + e^{-2x-0}} $$

$$ \lim_{h \to 0} \dfrac{1}{1 + e^{-x} + e^{-x} + e^{-2x}} $$

$$ \lim_{h \to 0} \dfrac{1}{1 + 2e^{-x} + e^{-2x}} $$

Factoring the denominator

$$ \lim_{h \to 0} \dfrac{1}{(1 + e^{-x})^2} $$

Conclusion:

$$ =\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{h} \cdot\dfrac{1}{(1 + e^{-(x+h)})\cdot (1 + e^{-x})} $$

$$ =\lim_{h \to 0} \dfrac{e^{-x}}{1} \cdot \dfrac{1}{(1 + e^{-x})^2} $$

$$ \lim_{h \to 0} \dfrac{\dfrac{1}{1 + e^{-(x+h)}} - \dfrac{1}{1 + e^{-x}}}{h} = \dfrac{e^{-x}}{(1 + e^{-x})^2} $$

rafaelcb21
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