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ET外星人 寫到:

f(x)=e<SUP>x</SUP>/(1+e<SUP>x</SUP>)

f'(x)=(e<SUP>x</SUP>(1+e<SUP>x</SUP>)-e<SUP>x</SUP>e<SUP>x</SUP>)/(1+e<SUP>x</SUP>)<SUP>2</SUP>

<SUP>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </SUP>=1/(1+e<SUP>x</SUP>)<SUP>2</SUP>

&nbsp;

f(x)=(3-x)/(x ln x)

f'(x)=(-(x ln x)-(3-x)(x(1/x)+ln x))/(x ln x)<SUP>2</SUP>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =(-x ln x -(3-x+3ln x -x ln x))/(x ln x)<SUP>2</SUP>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =(-3+x-3ln x)/(x ln x)<SUP>2</SUP>

&nbsp;

f(x)=ln ((x<SUP>2</SUP>-4)/(e<SUP>x</SUP>+1))

f'(x)=((2x(e<SUP>x</SUP>+1)-(x<SUP>2</SUP>-4)e<SUP>x</SUP>)/(e<SUP>x</SUP>+1)<SUP>2</SUP>)/((x<SUP>2</SUP>-4)/(e<SUP>x</SUP>+1))

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =((2x(e<SUP>x</SUP>+1)-(x<SUP>2</SUP>-4)e<SUP>x</SUP>)/(x<SUP>2</SUP>-4)(e<SUP>x</SUP>+1)

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =2x/(x<SUP>2</SUP>-4)-e<SUP>x</SUP>/(e<SUP>x</SUP>+1)

&nbsp;

不知對不對

如有錯請指教

第一題解
f'(x)= e^x/(e^x+1) ^2


第三題另解
f(x)=ln(x^2-4)-ln(e^x+1)
f'(x)=2x/(x^2-4)- e^x/(e^x+1)


琦琦的答案看不到.

第一題的答案



第二題的答案





第三題答案

#ed_op#DIV#ed_cl#f(x)=e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#/(1+e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f'(x)=(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#(1+e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)-e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)/(1+e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl#&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #ed_op#/SUP#ed_cl#=1/(1+e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(x)=(3-x)/(x ln x)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f'(x)=(-(x ln x)-(3-x)(x(1/x)+ln x))/(x ln x)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =(-x ln x -(3-x+3ln x -x ln x))/(x ln x)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =(-3+x-3ln x)/(x ln x)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(x)=ln ((x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4)/(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1))#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f'(x)=((2x(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1)-(x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4)e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)/(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)/((x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4)/(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1))#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =((2x(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1)-(x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4)e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#)/(x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4)(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =2x/(x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4)-e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#/(e#ed_op#SUP#ed_cl#x#ed_op#/SUP#ed_cl#+1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#不知對不對#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#如有錯請指教#ed_op#/DIV#ed_cl#

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