f==sin(2*x^2)/2
f'=2*x*cos(2*x^2);
f'(根號下pai/x）=(2*pi*cos((2*pi^2)/x^2))/x沒看懂.....不是吧！汗~f(x)=定積分上限為x^2,積分結(jié)果：f=sin(2*x^2)/2；對f求導(dǎo)：f'=2*x*cos(2*x^2);要求f'(根號下pai/x）=？將根號下pai/x代入f‘表達(dá)式，結(jié)果：f'(根號下pai/x）=(2*pi*cos((2*pi^2)/x^2))/x
f(x)=定積分上限為x^2,下限為0的cos2tdt,則f'(根號下pai/x）=
f(x)=定積分上限為x^2,下限為0的cos2tdt,則f'(根號下pai/x）=
數(shù)學(xué)人氣：225 ℃時(shí)間：2020-07-28 02:27:33
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