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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"微分の応用"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"X(t)に関して\n速度dx/dt=vとする。…①\nすると、加速度d^2x/dt^2=d/dt•(dx/dt)=dv/dt …②\nとなる。\n次にt(x)に関して\ndt/dx=1/(dx/dt)=(①を用いて)=1/v…③であり、\nd^2t/dx^2=d/dx•(dt/dx)=(③を用いて)=d/dx•(1/v)\n(これは合成関数の微分に相当するので)\n=-1/v^2•dv/dx=(vの変数としてのxはかなり扱いづらいので、tに変数変換して)=-1/v^2•dv/dt•dt/dx\nとなる。②、③を用いて変形すると、\nd^2x/dt^2=-v^3•d^2t/dx^2\nとなる。あとは①を代入して、答えは\n{}=-(dx/dt)^3となります。\n\nあってるかな、、？なんにせよこうゆうのにチャレンジしてみる姿勢は素晴らしいと思います。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"対数の計算"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"そもそも対数：logとは何でしょうか。\n　a^p＝M（a＞0, a≠1, M＞0）･･･①\nという等式が成り立つとき、\n　log a（M）＝p･･･②\nという等式が同時に成り立ちます。aを「底」、pを「指数」、Mを「真数」といい、log a（M）を「aを底とするMの対数」といいます。式②を見ればわかるように、log a（M）とは、「aを何乗したときMになるか」と言う値、すなわち、指数を表すものです。例えば、\n　log a（XY）＝log a（X）＋log a（Y）･･･③\nが成り立つのも、このとき\n　a^s＝XY\nと言う関係が常に存在し、X＝a^t、Y＝a^uとすると、\n　XY（＝a^s）＝X × Y\n　　　　　　　＝a^t × a^u\n　　　　　　　＝a^（t＋u）\nとなり、したがって、\n　s＝t＋u･･･④\nという関係を導くことができるからです。①と②から、XY、X、Yについても同様に、\n　log a（XY）＝s＝t＋u\n　log a（X）＝t\n　log a（Y）＝u\nと表せるので、結果として④は、\n　log a（XY）＝log a（X）＋log a（Ｙ）\nという式③になります。このように、logは、「対数」という名はあれど、その実「指数」のことを表しているのだということを頭に置いておくこと、つまり、①と②の対応関係を常に意識することが対数の理解の一助になるかもしれません。「logの数の大きい問題」というのがどんな問題を指すのかわからなかったので、ご期待に沿う回答ではないかもしれませんが、ご容赦ください。また、私の理解が誤っている場合は、これも申し訳ございません。"}],["$","div",null,{"className":"flex 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