Derivitives of exponential functions question

[wind]

Hi, I just have a quick question

Differentiate

\(\displaystyle \L\ h(x)=\frac{e^{-x}}{1-e^{-2x}}\)



\(\displaystyle \L\ f(x)=e^{-x}\).............\(\displaystyle \L\ f'(x)=e^{-x}*lne*1\)

\(\displaystyle \L\ g(x)= 1-e^{-2x}\)......\(\displaystyle \L\ g'(x)=-e^{-2x}*lne*-2\)


I'm not sure about this, if g'(x) has a negative e, then when you mulitply it by "ln base", would that include the negative( so ln-e)? But you can't take the log of a negative number...so does that mean that something is wrong?

Thanks

 

[galactus]

We can do the quotient rule thing.

\(\displaystyle \L\\\frac{(1-e^{-2x})(-e^{-x})-(e^{-x})(2e^{-2x})}{(1-e^{-2x})^{2}}\)

=\(\displaystyle \L\\\frac{-e^{x}(e^{2x}+1)}{(e^{2x}-1)^{2}}\)

 

[soroban]

Hello, wind!

Differentiate: \(\displaystyle \L\:h(x)\:=\:\frac{e^{^{-x}}}{1\,-\,e^{^{-2x}}}\)

Those negative exponents are very annoying . . .

Multiply by \(\displaystyle \L\frac{e^{^{2x}}}{e^{^{2x}}}:\)

. . \(\displaystyle \L h(x) \:=\: \frac{e^{^{2x}}}{e^{^{2x}}}\,\cdot\,\frac{e^{^{-x}}}{1\,-\,e^{^{-2x}}} \;=\;\frac{e^{^x}}{e^{^{2x}}\,-\,1}\)


Differentiate (Quotient Rule):

. . \(\displaystyle \L h'(x)\;=\;\frac{(e^{^{2x}}\,-\,1)\cdot e^{^x}\,-\,e^{^x}(2e^{^{2x}})} {(e^{^{2x}}\,-\,1)^2} \;=\;\frac{e^{^x}\left[(e^{^{2x}}\,-\,1)\,-\,2e^{^x}\right]}{(e^{^{2x}}\,-\,1)^2}\)

. . . . . . . . \(\displaystyle \L=\;\frac{e^{^x}(-e^{^{2x}}\,-\,1)}{(e^{^{2x}}\,-\,1)^2} \;=\;-\frac{e^{^x}(e^{^{2x}}\,+\,1)}{(e^{^{2x}}\,-\,1)^2}\)