Original problem (third problem here)
Plane $T$ nears the surface $S$ $S: \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt = \ln(2)$ in a point that is on positive $z$ -axis. Assign $T$'s equation.
So I think I need $\nabla S\times T=0$ where $T=\begin{pmatrix}x \\ y \\ z \end{pmatrix}$. So
$\partial_x S := \partial_x \left( \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt \right)=?$
Now should I integrate this before differentiation or was there some rule to make this differentation-integration simpler?
Trial 4 By Leibniz rule kindly suggested by the engineer and the formula by Petersen,
$\begin{align}\partial_x \int_0^1 f(x,t) dt \Bigg\vert_{x=0} &= \int_0^1 \partial_x f(x,t) \Big\vert_{x=0} dt \\ &=z(y-z)\log\left(\frac{1-y+z}{-y+z}\right)+z+\frac{1}{-y+z+1}-\frac{1}{-y+z} \\ &:=B \end{align}$
Details here. Now after this, the same for $\partial_y S:=C$ and $\partial_z S :=D$ so $\nabla S = \begin{pmatrix} B \\ C \\ D\end{pmatrix}$.
But to the question, is $\nabla S = \begin{pmatrix} B \\ ... \\ ...\end{pmatrix}$ right? Or $\nabla S_{|\bar{x}=(0,0,1)} = \begin{pmatrix} -\log(2)+0.5 \\ ... \\ ...\end{pmatrix}$? Look after this I need to do $\nabla S\times T=0$ and I want to minimize the amount of terms early on because it is easy to do mistakes with long monotonous calculations.
Old trials in the chronological order
T2: Plane and surface meeting at the point (0,0,1), deduction with "extra-minus" mistake
T3: trying to calculate the $\nabla$ (this uses wrong point (0,0,-2) instead of (0,0,1) but is the idea correct?) (here)