What is the derivative of $\int_{-10}^{-3} e^{\tan(t)} \,dt$ with respect to x?

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We were learning about the Fundamental Theorem of Calculus today in my high school and the above integral came up as an example of an integral with a "constant" value. At first I accepted that the derivative of it was zero, but then I realized that since derivatives are defined for radian versions of the trigonometric functions, $e^{\tan(t)}$ does not satisfy the continuity requirement for it to be guaranteed to have an convergent integral on the interval $[-10, -3]$.

I had my suspicions that the integral did not exist as a real number, so I checked it with the NINT tool on my graphing calculator and got $\infty$. I also cross-validated the result with WolframAlpha and reaffirmed that the integral does not converge to any real number.

Yet a peer pointed out that the calculator also evaluated the derivative of $\infty$ as $0$, and WolframAlpha confirms that as well.

My Question

Is the derivative of $\infty$ truly $0$? It seems rediculous to me to treat it as a constant since it's only a concept, but has someone defined a use or reason for it to be that way?

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Zero. The value of the integral doesn't depend on $x$ at all.