Natural Logs

[thelazyman]

Can someone tell me if e^-x derives to -e^-x


And can someone please tell me the derivative of xe^-x


Thanks

 

[Unco]

Can someone tell me if e^-x derives to -e^-x

It certainly does.

And can someone please tell me the derivative of xe^-x

Apply the product rule:

\(\displaystyle \mbox{ \frac{d}{dx} uv = u\cdot\frac{dv}{dx} + v\cdot\frac{du}{dx}}\)

If \(\displaystyle \mbox{u=x}\) and \(\displaystyle \mbox{v=e^{-x}}\), we have

\(\displaystyle \begin{align*}
\mbox{ \frac{d}{dx} x\cdot e^{-x}} &= \mbox{x \cdot (-e^{-x}) + e^{-x} \cdot 1} \\
\\
\mbox{ } &= \mbox{- x\cdot e^{-x} + e^{-x} } \\
\end{align*}\)

 

[Unco]

If you wish to use natural logs,

let \(\displaystyle \mbox{ y = xe^{-x}}\)

Take natural logs of both sides:

\(\displaystyle \mbox{ \ln{y} = \ln{(xe^{-x}})}\)

Use the log of a product rule:

\(\displaystyle \mbox{ \ln{y} = \ln{x} + \ln{(e^{-x})}\)

Use the log of a power rule:

\(\displaystyle \mbox{ \ln{y} = \ln{x} -x\ln{e}\)

And \(\displaystyle \mbox{ \ln{e} = 1}\):

\(\displaystyle \mbox{ \ln{y} = \ln{x} -x}\)

Implicitly differentiate both sides with respect to \(\displaystyle \mbox{x}\):

\(\displaystyle \mbox{ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - 1}\)

\(\displaystyle \mbox{ \frac{dy}{dx} = y(\frac{1}{x} - 1)}\)

But \(\displaystyle \mbox{y = xe^{-x}}\). Substitute and you get the same answer as above.