Fourier transform rules for Dirac-deltas, to avoid integration I'm following through a worked example but have a different solution to the given solution. Could someone tell me what I'm missing please, if anything?  
I have taken the 2nd derivative of a piece-wise function, ending up with:  
$$\frac{d^2f(t)}{dt^2}=
\frac {d}{dt}\delta \left( t+T \right)
-\frac {d}{dt}\delta \left( t-T \right)
+\frac {1}{T}\delta \left( t+T \right)
+\frac {1}{T}\delta \left( t-T \right)
$$
This matches with the solution I have so far.  
My solution to find the Fourier transform is 
$$\mathcal F\left[\frac{d^2f(t)}{dt^2}\right]=j\omega\left(e^{j\omega T}-e^{-j\omega T}\right)+\frac{1}{T}\left(e^{j\omega T}+e^{-j\omega T}\right)$$
which is simplified to 
$$-2\omega\sin(\omega T)+\frac{2}{T}\cos(\omega T)$$
The solution I have been given states (though it may have a typo, and be $-2/T$): 
$$-2\omega\sin(\omega T)+\frac{2}{T}\cos(\omega T)+\frac{2}{T}$$
I'm not sure what I've missed to get the $+\frac{2}{T}$. 
I have used:
$$\mathcal F[\delta(t)]=1$$ and that the derivative property of: 
$$\mathcal F\left[\frac{d\delta(t)}{d t}\right]=j\omega$$
and the time-shift property of:
$$\mathcal F \left[\delta \left( t+T \right)\right]=e^{j\omega T}$$
Can someone explain what I missed please? I'm thinking it might be when the time shifted Dirac delta is differentiated, but I'm not sure how it works and can't find any literature.  
(For completion, the final step of the main problem, not included here, is to integrate twice  by multiplying by $\frac{1}{(j\omega)^2}$ )
 A: It is helpful to split the function of interest
$$f(t)=\left\{ \begin{array}{cc} 2-|t|/T, & -T<t<T \\ 0, & \text{else}\end{array} \right.$$ as $f(t)=f_1(t)+f_2(t)$ where 
$$f_1(t)=\left\{
\begin{array}{cc} 2-|t|/T, & -T<t<T \\ 0, & \text{else}\end{array}\right.,
\qquad f_2(t)=
\left\{ \begin{array}{cc} 1-|t|/T, & -T<t<T \\ 0, & \text{else} \end{array}\right.$$
Note that $f_1(t)$ is discontinuous but locally constant and $f_2(t)$ is continuous but not smooth.
We now consider the two pieces separately. For $f_1(t)$ we may write 
$$f_1(t) = \theta(t+T)-\theta(t-T)$$ where $\theta(t)$ is the Heaviside step function. Hence $$\frac{df_1}{dt} = \delta(t+T)-\delta(t-T)$$ and therefore $$\mathcal{F}\left[\frac{df_1}{dt}\right] = e^{j \omega T}-e^{-j \omega T}=2j\sin\omega T.$$ The derivative rule then yields $$\mathcal{F}\left[\frac{d^2f_1}{dt^2}\right]=(j\omega) \mathcal{F}\left[\frac{d^2f_1}{dt^2}\right] = -2 \omega \sin(\omega T).$$ This agrees with the work in the OP, so the locally-constant part was handled correctly.
What about the continuous-but-not-smooth part? Writing this part as $$f_2(t) = \left\{ \begin{array}{cc} 1+t/T, & -T<t<0 \\ 1-t/T, & 0<t<T \\ 0, & \text{else} \end{array}\right.,$$ we have $$\frac{df_2}{dt} = \left\{ \begin{array}{cc} 1/T, & -T<t<0 \\ -1/T, & 0<t<T \\ 0, & \text{else} \end{array}\right.$$ This is locally constant with three discontinuities, so its expression in terms of Heaviside step functions requires three terms: $$\frac{df_2}{dt} = \frac{1}{T}\theta(t+T)-\frac{2}{T}\theta(t)+\frac{1}{T}\theta(t-T).$$ Thus $$\frac{d^2 f_2}{dt^2} = \frac{1}{T}\delta(t+T)-\frac{2}{T}\delta(t)+\frac{1}{T}\delta(t-T)l,$$ which differs from the OP due to the second term. 
Therefore 
$$\mathcal{F}\left[\frac{df_2}{dt}\right] = \frac{1}{T}e^{j\omega T}-\frac{2}{T}(1)+\frac{1}{T}e^{-j\omega T}=\frac{2}{T}\cos(\omega T)-\frac{2}{T}.$$ The extra term $-2/T$ thus reflects the kink in $f_2(t)$ at $t=0$, which $f(t)$ of course also possesses.
As an alternative to the above splitting, one may write $f(t)$ itself in terms of Heaviside step functions: 
\begin{align}
f(t)
&= [\theta(t+T)-\theta(t)](1+t/T)+[\theta(t)-\theta(t-T)](1-t/T) \\
&= \left(1+\frac{t}{T}\right)\theta(t+T)-\frac{2}{T}\theta(t)+\left(1-\frac{t}{T}\right)\theta(t-T).
\end{align}
The reader is invited to verify that differentiating this expression twice yields the same result as above.
