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Question Number 64327 by Chi Mes Try last updated on 16/Jul/19

(d/dx)  ( ∫_(f(x)) ^(g(x))  φ(t) dt ) =

$$\frac{{d}}{{dx}}\:\:\left(\:\underset{{f}\left({x}\right)} {\overset{{g}\left({x}\right)} {\int}}\:\phi\left({t}\right)\:{dt}\:\right)\:= \\ $$

Commented by mathmax by abdo last updated on 16/Jul/19

(d/dx)( ∫_(f(x)) ^(g(x)) ϕ(t)dt ) =g^′ (x)ϕ(g(x))−f^′ (x)ϕ(f(x))  example  (d/dx)(  ∫_x ^x^2    t(√(t+1))dt) =(2x)x^2 (√(x^2  +1))−x(√(x+1)).

$$\frac{{d}}{{dx}}\left(\:\int_{{f}\left({x}\right)} ^{{g}\left({x}\right)} \varphi\left({t}\right){dt}\:\right)\:={g}^{'} \left({x}\right)\varphi\left({g}\left({x}\right)\right)−{f}^{'} \left({x}\right)\varphi\left({f}\left({x}\right)\right) \\ $$$${example}\:\:\frac{{d}}{{dx}}\left(\:\:\int_{{x}} ^{{x}^{\mathrm{2}} } \:\:{t}\sqrt{{t}+\mathrm{1}}{dt}\right)\:=\left(\mathrm{2}{x}\right){x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}−{x}\sqrt{{x}+\mathrm{1}}. \\ $$

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