Question Number 224733 by fantastic last updated on 30/Sep/25
$${Let}\:{f}\:{be}\:{a}\:{continuously}\:{differentiable}\:{function} \\ $$$${such}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{x}^{\mathrm{2}} } {f}\left({t}\right){dt}={e}^{\mathrm{cos}\:{x}^{\mathrm{2}} } \:{for}\:{all}\:{x}\in\left(\mathrm{0},\infty\right) \\ $$$${the}\:{value}\:{of}\:{f}\:'\left(\pi\right)=? \\ $$
Answered by mr W last updated on 01/Oct/25
$${f}\left(\mathrm{2}{x}^{\mathrm{2}} \right)×\mathrm{4}{x}=−\mathrm{2}{x}\mathrm{sin}\:{x}^{\mathrm{2}} {e}^{\mathrm{cos}\:{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{f}\left(\mathrm{2}{x}^{\mathrm{2}} \right)=−\mathrm{sin}\:{x}^{\mathrm{2}} {e}^{\mathrm{cos}\:{x}^{\mathrm{2}} } \\ $$$${let}\:{t}=\mathrm{2}{x}^{\mathrm{2}} \\ $$$${f}\left({t}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{t}}{\mathrm{2}}{e}^{\mathrm{cos}\:\frac{{t}}{\mathrm{2}}} \\ $$$${f}'\left({t}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}−\mathrm{cos}\:\frac{{t}}{\mathrm{2}}\right){e}^{\mathrm{cos}\:\frac{{t}}{\mathrm{2}}} \\ $$$${f}'\left(\pi\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{2}}−\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\right){e}^{\mathrm{cos}\:\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by fantastic last updated on 01/Oct/25
$${you}\:{are}\:{right} \\ $$