$$\frac{d}{dt}\int_a^b f(x,t)\,dx=\int_a^b \partial_t f(x,t)\,dx$$

Key takeaway: This identity is a standard differentiation-under-the-integral-sign result when the bounds are constant. Since the limits do not depend on $t$, the derivative passes through the integral and acts only on the integrand.

Because the limits of integration are constant, the variable $t$ appears only inside the integrand. Under the usual continuity and differentiability assumptions on $f(x,t)$, differentiation with respect to $t$ can be moved inside the integral:

$\frac{d}{dt}\int_a^b f(x,t)\,dx=\int_a^b \partial_t f(x,t)\,dx.$

A clean way to see this is through the definition of the derivative:

=\lim_{h\to 0}\frac{1}{h}\left(\int_a^b f(x,t+h)\,dx-\int_a^b f(x,t)\,dx\right).$$ Combine the integrals: $$=\lim_{h\to 0}\int_a^b \frac{f(x,t+h)-f(x,t)}{h}\,dx.$$ If $\partial_t f$ is continuous, the quotient converges to $\partial_t f(x,t)$, so the limit becomes $$\int_a^b \partial_t f(x,t)\,dx.$$ That gives the stated formula. --- **Pitfalls the pros know** 👇 The main mistake is confusing this rule with the Leibniz rule for moving endpoints. Here the limits $a$ and $b$ are constants, so there are no boundary terms. Another common error is forgetting that the derivative is with respect to $t$, not $x$. **What if the problem changes?** If the limits also depended on $t$, for example $\int_{a(t)}^{b(t)} f(x,t)\,dx$, then the derivative would include boundary terms: $$\frac{d}{dt}\int_{a(t)}^{b(t)} f(x,t)\,dx=\int_{a(t)}^{b(t)} \partial_t f(x,t)\,dx+f(b(t),t)b'(t)-f(a(t),t)a'(t).$$ `Tags`: differentiation under the integral sign, Leibniz rule, partial derivative