Integration by Parts and Leibniz Rule for Differentiation under the Integral Sign Basically a friend of mine and I have had this hot debate for a little too long, I contend that these two tools are not only logically unconnected but they require different assumptions (I believe one requires a continuously differentiable function and another requires it to simply be continuous). We've even gone through the proofs and disagree on how the assumptions are used. I don't see the connection...  Maybe I'm wrong, maybe they are equivalent (you have one as a tool if and only if you have the other). Anyway, any fresh perspective would be welcomed and any deeper discussion on either appreciated, thanks.
 A: Okay! So I think I have an answer to my own question and I would appreciate it if I had some confirmation on this.
Define $f(x)$ to be an continuously differentiable function, and $g(x)$ to be the Weierstrass function.
$$g(x)=\sum_{n=0}^{\infty}a^n\cos(b^n\pi x) : 0 <a <1, b\in2\mathbb{Z}, ab>1+\frac{3}{2}\pi $$
Clearly integration by parts is not well defined because 
$$[f(x)g(x)]|^\beta_\alpha = \int_\alpha^\beta f' (x)g(x)dx+ \int_\alpha^\beta f(x){\color{red}{g'(x)}}dx $$
this statement does not make sense as $g$ is not differentiable anywhere.  
On the other hand, we may come up with an interesting case where the Leibniz rule applies; namely, if we define the convolution as
$$(f\star g)(t) = \int_{\tau=0}^{\tau=t} g(\tau)f(t-\tau)d\tau$$
Then in particular we may show 
$$\frac{d}{dx}(f\star g)(x)=g(0)f(x)+(f'\star g)(x)$$
using the Leibniz rule for differentiating under the integral sign.
Finally, in our case if we apply the convolution we see that 
$$\frac{d}{dx} (f\star g)(x) = \sum_{n=0}^\infty a^n f(x) + (f'\star g)(x)$$
which is well defined because $f$ is continuously differentiable and the non-differentiability of $g$ is not a problem.
