derivaties with e/ simple question

[delgeezee]

help to find the derivative of
\(\displaystyle f(x)=xe^{-x}\)
i will use product rule

My question is if the derivative of \(\displaystyle e^{-x} = -xe^{-x} \)???

 

[tkhunny]

Where did you get the spare 'x'?

\(\displaystyle \frac{d}{dx}e^{x} = e^{x} \ne x\cdot e^{x}\)

\(\displaystyle \frac{d}{dx}e^{-x} = -e^{-x} \ne -x\cdot e^{-x}\)

 

[renegade05]

hello delgeezee!

You must use the chain rule and product rule for this one.

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

Think of the \(\displaystyle -x\) as the "inner function". So after you take the derivative of \(\displaystyle e^{-x}\) which is just \(\displaystyle e^{-x}\) you must then take the derivative of \(\displaystyle -x\) which is \(\displaystyle -1\) hence: \(\displaystyle \frac{d}{dx}(e^{-x})=-e^{-x}\).

So applying this to your original problem we have:

\(\displaystyle \frac{d}{dx}(xe^{-x})=\underbrace{\frac{d}{dx}(x)e^{-x}+x\frac{d}{dx}(e^{-x})}_{Product~rule}=e^{-x}\underbrace{-xe^{-x}}_{Chain~rule}\)

\(\displaystyle f'(x)=e^{-x}-xe^{-x}\)

 

[delgeezee]

hello delgeezee!

You must use the chain rule and product rule for this one.

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

Think of the \(\displaystyle -x\) as the "inner function". So after you take the derivative of \(\displaystyle e^{-x}\) which is just \(\displaystyle e^{-x}\) you must then take the derivative of \(\displaystyle -x\) which is \(\displaystyle -1\) hence: \(\displaystyle \frac{d}{dx}(e^{-x})=-e^{-x}\).

So applying this to your original problem we have:

\(\displaystyle \frac{d}{dx}(xe^{-x})=\underbrace{\frac{d}{dx}(x)e^{-x}+x\frac{d}{dx}(e^{-x})}_{Product~rule}=e^{-x}\underbrace{-xe^{-x}}_{Chain~rule}\)

\(\displaystyle f'(x)=e^{-x}-xe^{-x}\)

Thank you renegade. Your explanation recalled my memory.