Mathematics
Senior High
Resolved
第二次導関数の解き方教えてください
第二
y'=2e-2x{(b-a)cos2x-(a+b)sin 2x }
y'=8e2*(-bcos2x + asin2x
よって
Answers
Answers
Were you able to resolve your confusion?
Users viewing this question
are also looking at these questions 😉
Recommended
詳説【数学Ⅰ】第二章 ２次関数（後半）～最大・最小・不等式～
6135
25
詳説【数学Ⅰ】第二章 ２次関数（前半）～関数とグラフ～
5157
18
詳説【数学Ａ】第３章 平面図形
3626
16
詳説【数学Ⅱ】第２章 図形と方程式（上）～点と直線～
2685
13