Question Number 17480 by Arnab Maiti last updated on 06/Jul/17
$$\mathrm{why}\:\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\int_{\mathrm{0}} ^{\:\:\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}\right)=\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Answered by mrW1 last updated on 06/Jul/17
![let F(y)=∫_0 ^y e^t dt=[e^t ]_0 ^y =e^y −1 (dF/dy)=e^y ⇒(dF/dx)=(dF/dy)×(dy/dx)=e^y (dy/dx)](Q17483.png)
$$\mathrm{let}\:\mathrm{F}\left(\mathrm{y}\right)=\int_{\mathrm{0}} ^{\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}=\left[\mathrm{e}^{\mathrm{t}} \right]_{\mathrm{0}} ^{\mathrm{y}} =\mathrm{e}^{\mathrm{y}} −\mathrm{1} \\ $$$$\frac{\mathrm{dF}}{\mathrm{dy}}=\mathrm{e}^{\mathrm{y}} \\ $$$$\Rightarrow\frac{\mathrm{dF}}{\mathrm{dx}}=\frac{\mathrm{dF}}{\mathrm{dy}}×\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Commented by Arnab Maiti last updated on 06/Jul/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Now}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{well}. \\ $$