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Suppose $q(t)=A/(t-c)+r(t)$, where $r$ is continuous, the pole $c$ is fixed, and differentiable endpoints obey $a(x)<c<b(x)$ locally. ThenApply the chain rule and the fundamental theorem of calculus to getThe same boundary expression as in the Leibniz integral rule thus applies to this explicitly specified Cauchy principal value. If $A\ne 0$, the two ordinary one-sided improper integrals diverge, so their sum is not an ordinary integral. A finite boundary derivative alone does not make that divergent integral meaningful.

The Leibniz integral rule accounts for both the moving upper endpoint and the integrand's dependence on $a$:For the specified integrand this gives $I^{\prime }(a)=e^{-a^2}-{\int }_0^{a}x{e}^{-ax}dx$. When $a\ne 0$, integration by parts givessoFor the requested independent check, direct integration gives $I(a)=(1-e^{-a^2})/a$ at $a\ne 0$. Its derivative is $[2a^2{e}^{-a^2}-(1-e^{-a^2})]/a^2$, agreeing with the result above. At zero the apparent singularity is removable: $I(a)=a-a^3/2+O(a^5)$, so $\boxed{I(0)=0,I^{\prime }(0)=1}$.

Under the regularity and integrability conditions for the Leibniz integral rule,The first two terms account for the moving endpoints; the last accounts for the integrand's own parameter dependence.

The integrand and both limits depend on $x$, so use the Leibniz integral rule. The upper and lower endpoint contributions and the interior derivative giveFor $x\ne 0$, an antiderivative in $t$ is $t\sin (xt)/x+\cos (xt)/x^2$. Evaluating it and collecting terms givesAt $x=0$ the original Leibniz expression has zero endpoint terms and an integral over a zero-length interval, giving $\boxed{F^{\prime }(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.

Apply the Leibniz integral rule, including both moving boundaries and the explicit parameter in the integrand. The upper bound has derivative $-\sin x$, the lower bound has derivative $\cos x$, and the integrand has partial derivative $-t^4{e}^{-x{t}^4}$ with respect to $x$. ThereforeThe formula remains valid when the bounds coincide or occur in reversed order, by the oriented definite integral convention. The integrand and its partial derivative are continuous on every finite integration interval.

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

Under the usual regularity assumptions, the Leibniz integral rule isThere is a genuine domain problem in the following integral as printed in the PDF. PutThe point $t=2$ is a simple pole, withFor every $x>4$, the lower endpoint $\sin (1/x)$ is less than $2$ and the upper endpoint $\sqrt{x}$ is greater than $2$. The interval therefore crosses the pole. Since $q(t)=33/[7(t-2)]+O(1)$ near $2$, the two one-sided improper integrals diverge logarithmically, with opposite signs. Their divergences do not define an ordinary integral. Consequently the ordinary limit in the question is undefined as written: its underlying integral has no finite value for any $x\ge 4$.

A separately specified Cauchy principal value does exist for $x>4$. Under that interpretation, differentiating a principal-value integral across a fixed simple pole givesThe first term isand the second tends to zero because $q$ is bounded near $t=0$. Thus the principal-value version has limit $\boxed{1}$. The two conclusions are distinct: this regularised answer requires the additional principal-value convention, which the printed question does not state.

The Leibniz integral rule combines contributions from both moving endpoints with differentiation under the integral sign:The partial derivative in the last integral holds $t$ fixed. Suitable regularity and convergence are assumed as stipulated.

For the Leibniz integral rule, the upper endpoint has derivative $2\pi x$ and the lower endpoint has derivative $1$. The integrand has partial derivative $4xt+\sin t$. ThereforeThe last integral is $[2x{t}^2-\cos t{]}_x^{\pi x^2}=2{\pi }^2{x}^5-2x^3-\cos (\pi x^2)+\cos x$. Combining terms gives