I'm working through a Green's function problem for a second-order linear differential operator $L$ and I need to find the derivative of: $$ u(x) = \frac{u_2(x)}{a_0(\xi)W(\xi)} \int_a^x u_1(\xi)f(\xi) \ d\xi \ + \frac{u_1(x)}{a_0(\xi)W(\xi)} \int_a^x u_2(\xi)f(\xi) \ d\xi $$ I am thinking this has to be done using Leibniz integral rule for variable bounds. Since $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(a)=0$ I think it should be $$ u'(x) = \frac{u_2'(x)}{a_0(\xi)W(\xi)}\int_a^xu_1(\xi)f(\xi) \ d\xi + \frac{u_1'(x)}{a_0(\xi)W(\xi)}\int_a^xu_2(\xi)f(\xi) \ d\xi $$ where the $f(x,b(x))\cdot\frac{d}{dx}b(x) $ term from the first part cancels out with the $-f(x,a(x))\cdot\frac{d}{dx}a(x) $ of the second part. Is this correct? It comes out so simple in the end, I am also wondering if there was an easier way to come to this conclusion.
Additionally, when we take the second derivative will we see that term cancel out again? Or will it not because in the second derivative we will have $u_2'(x)u_1(x)f(x)-u_1'(x)u_2(x)f(x)$? Thanks for any help!
By the Leibniz integral rule, you would have, for instance,
$$\begin{align} \frac{\mathrm d}{\mathrm dx}\left[\int_a^xu_2(x)u_1(\xi)f(\xi)\,\mathrm d\xi\right]&=u_2(x)u_1(x)f(x)\cdot\frac{\mathrm d(x)}{\mathrm dx}-u_2(x)u_1(a)f(a)\cdot\frac{\mathrm d(a)}{\mathrm dx}\\ &\quad\quad+\int_a^x\frac{\partial}{\partial x}\left[u_2(x)u_1(\xi)f(\xi)\right]\,\mathrm d\xi\\[1ex] &=\underbrace{u_2(x)u_1(x)f(x)}{}+u_2'(x)\int_a^xu_1(\xi)f(\xi)\,\mathrm d\xi \end{align}$$