Could someone please help me with these derivatives?
I have to find the first derivative of
$$f(x) = \frac{e^{-2x}}{1-x}.$$
Then I have to find the second derivative of that.
For the first derivative I get
$$ f'(x) = \frac{-2e^{-2x}}{(1-x)^2} $$
The answer book disagrees with me though. What am I doing wrong?
$\endgroup$ 02 Answers
$\begingroup$$\left(\dfrac{h}{g}\right)'=\dfrac{h'g-hg'}{g^2}$
You have $h(x)=e^{-2x},\,h'(x)=-2e^{-2x},\,g(x)=1-x,\,g'(x)=-1$.
$\endgroup$ 0 $\begingroup$Let $\displaystyle f(x)=\frac{u(x)}{v(x)}$, where $u=e^{-2x} \Rightarrow u'=-2e^{-2x} $ and $v=1-x \Rightarrow v'=-1$.
Now, as $\displaystyle\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}$, then $$f'(x) =\frac{(1-x)(-2e^{-2x})-e^{-2x}(-1)}{(1-x)^2}=\frac{-2e^{-2x}+2xe^{-2x}+e^{-2x}}{(1-x)^2}=\frac{(2x-1)e^{-2x}}{(1-x)^2}.$$
$\endgroup$ 3
