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How can I perform derivative of double integral

$$\frac{\mathrm d}{\mathrm dt}\int_{t-\mathrm d1}^t \int_h^t f(s) \,\mathrm ds\,\mathrm dh$$

Can I apply a Leibniz rule of some form? How?

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  • $\begingroup$ Do you know what the answer is supposed to be? What have you tried? $\endgroup$ – Rumplestillskin Jan 18 '17 at 6:17
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Let $\int_h^tf(x)ds = F(t)-F(h)$ with $F'(t)=f(t)$. Then

\begin{align} \frac{\mathrm d}{\mathrm dt}\int_{\color{red}{t-\mathrm d_1}}^{\color{blue}{t}} \int_h^t f(s) \,\mathrm ds\,\mathrm dh&=\frac{\mathrm d}{\mathrm dt}\int_{\color{red}{t-\mathrm d_1}}^{\color{blue}{t}} (F(t)-F(h))\,\mathrm dh\\ &=\int_{\color{red}{t-\mathrm d_1}}^{\color{blue}{t}} \frac{\mathrm d}{\mathrm dt}(F(t)-F(h))\,\mathrm dh\\ &+\frac{\mathrm d}{\mathrm dt}(\color{blue}{t})\times (F(\color{blue}{t})-F(h))\\ &-\frac{\mathrm d}{\mathrm dt}(\color{red}{t-d_1})\times (F(\color{red}{t-d_1})-F(h))\\ &=\int_{\color{red}{t-\mathrm d1}}^{\color{blue}{t}} f(t)\,\mathrm dh+F(\color{blue}{t})-F(h)-F(\color{red}{t-d_1})+F(h)\\ &=d_1 f(t)+F(\color{blue}{t})-F(\color{red}{t-d_1})\\ &=d_1 f(t)+F(\color{blue}{t})-F(\color{red}{t-d_1})\\ \end{align} the latter could be simplified depending on the sign of $d_1$.

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HINT
Set $$\int_h^t f(s)\mathrm{d}s=F(t)$$ Then apply Leibniz twice

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