1 over e to the x Derivative Proof

[Jason76]

\(\displaystyle \dfrac{d}{dx} \dfrac{1}{e^{x}}\)

\(\displaystyle \dfrac{(e^{x})(0) - (1)(e^{x})}{(e^{x})^{2}}\) - Quotient Rule \(\displaystyle \dfrac{(g)(f') - (f)(g')}{g^{2}}\) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle \dfrac{-e^{x}}{?}\) What is \(\displaystyle (e^{x})^{2}\) ?

Note: The answer to the whole thing should be \(\displaystyle -e^{-x}\).

 

[DrPhil]

\(\displaystyle \dfrac{d}{dx} \dfrac{1}{e^{x}}\)

\(\displaystyle \dfrac{(e^{x})(0) - (1)(e^{x})}{(e^{x})^{2}}\) - Quotient Rule \(\displaystyle \dfrac{(g)(f') - (f)(g')}{g^{2}}\) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle \dfrac{-e^{x}}{?}\) What is \(\displaystyle (e^{x})^{2}\) ?

Note: The answer to the whole thing should be \(\displaystyle -e^{-x}\).

ALGEBRA.... \(\displaystyle \displaystyle (a^b)^c = a^{bc} = (a^c)^b\)

Never use quotient rule if you can use a negative power instead!!

\(\displaystyle \dfrac{1}{e^x} = e^{-x}\)

 

[Deleted member 4993]

\(\displaystyle \dfrac{d}{dx} \dfrac{1}{e^{x}}\)

\(\displaystyle \dfrac{(e^{x})(0) - (1)(e^{x})}{(e^{x})^{2}}\) - Quotient Rule \(\displaystyle \dfrac{(g)(f') - (f)(g')}{g^{2}}\) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle \dfrac{-e^{x}}{?}\) What is \(\displaystyle (e^{x})^{2}\) ?

Note: The answer to the whole thing should be \(\displaystyle -e^{-x}\).

Your knowledge of algebra is "biting" you:

the above is same as \(\displaystyle \dfrac{-a}{(a)^2}\) which is equal to \(\displaystyle -\dfrac{1}{a}\)

 

[Jason76]

ALGEBRA.... \(\displaystyle \displaystyle (a^b)^c = a^{bc} = (a^c)^b\)

Never use quotient rule if you can use a negative power instead!!

\(\displaystyle \dfrac{1}{e^x} = e^{-x}\)

\(\displaystyle \dfrac{1}{e^{x}} = e^{-x}\)

\(\displaystyle \dfrac{d}{dx} (e^{-x})\)

\(\displaystyle e^{u} du\)

\(\displaystyle e^{u} (-1)\)

\(\displaystyle -e^{u}\)

\(\displaystyle -e^{-x}\)

But it would still be interesting to see the result via quotient rule.

\(\displaystyle \dfrac{d}{dx} \dfrac{1}{e^{x}}\)

\(\displaystyle \dfrac{(e^{x})(0) - (1)(e^{x})}{(e^{x})^{2}}\) - Quotient Rule \(\displaystyle \dfrac{(g)(f') - (f)(g')}{g^{2}}\) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle -\dfrac{e^{x}}{(e^{x})^{2}}\)

\(\displaystyle -(e^{x})^{-1}\)

\(\displaystyle -e^{-x}\)

 

[Deleted member 4993]

But it would still be interesting to see the result via quotient rule.

It might be interesting to see somebody smashing a cone of ice-cream with mortar and pestle and then eat it...

But that would be a sheer waste of time and effort (like in this case).

 

[DrPhil]

It might be interesting to see somebody smashing a cone of ice-cream with mortar and pestle and then eat it...

But that would be a sheer waste of time and effort (like in this case).

More interesting (and perhaps even instructive) to see the limit process:

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

EDIT - comment - this is the fundamental way to PROVE any derivative

 

[Deleted member 4993]

More interesting (and perhaps even instructive) to see the limit process:

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

I agree ....