AP Calculusのmultiple choiceセクションは、計算機なしのパートが60分で30問だから1問平均2分しかありません。
計算機を使うパートは45分で15問だから1問平均3分ですが、計算機の操作の時間を含めるとやはり時間はかなり厳しいです。
いずれにしても基本的な公式を試験時間中に導き出している余裕はありません。この公式集で紹介する公式は覚えておきましょう。
ただし勉強する段階では導出も一緒に勉強しておくことが大事です。
微分法 Differentiation
$n$と$a$は定数、$u$と$v$は$x$で微分可能な関数とします。
\[\dfrac{da}{dx}=0\tag{1}\]
\[\dfrac{d}{dx}au=a\dfrac{du}{dx}\tag{2}\]
\[\dfrac{d}{dx}x^n=nx^{n-1}\qquad \text{(Power Rule)}\tag{3}\]
\[\dfrac{d}{dx}u^n=nu^{n-1}\dfrac{du}{dx}\tag{4}\]
\[\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}\tag{5}\]
\[\dfrac{d}{dx}(u-v)=\dfrac{du}{dx}-\dfrac{dv}{dx}\tag{6}\]
\[\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\qquad \text{(Product Rule)}\tag{7}\]
\[\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}\qquad (v\ne 0) \qquad \text{(Quotient Rule)}\tag{8}\]
\[\dfrac{d}{dx}\sin{u}=\cos{u}\times\dfrac{du}{dx}\tag{9}\]
\[\dfrac{d}{dx}\cos{u}=-\sin{u}\times\dfrac{du}{dx}\tag{10}\]
\[\dfrac{d}{dx}\tan{u}=\sec^2{u}\times\dfrac{du}{dx}\tag{11}\]
\[\dfrac{d}{dx}\cot{u}=-\csc^2{u}\times\dfrac{du}{dx}\tag{12}\]
\[\dfrac{d}{dx}\sec{u}=\sec{u}\tan{u}\times\dfrac{du}{dx}\tag{13}\]
\[\dfrac{d}{dx}\csc{u}=-\csc{u}\cot{u}\times\dfrac{du}{dx}\tag{14}\]
\[\dfrac{d}{dx}\ln{u}=\dfrac{1}{u}\times\dfrac{du}{dx}\tag{15}\]
\[\dfrac{d}{dx}e^{u}=e^{u}\times\dfrac{du}{dx}\tag{16}\]
\[\dfrac{d}{dx}a^u=a^u\ln{a}\times\dfrac{du}{dx}\tag{17}\]
\[\dfrac{d}{dx}\sin^{-1}{u}=\dfrac{1}{\sqrt{1-u^2}}\times\dfrac{du}{dx}\qquad (-1\lt u\lt 1)\tag{18}\]
\[\dfrac{d}{dx}\cos^{-1}{u}=-\dfrac{1}{\sqrt{1-u^2}}\times\dfrac{du}{dx}\qquad (-1\lt u\lt 1)\tag{19}\]
\[\dfrac{d}{dx}\tan^{-1}{u}=\dfrac{1}{1+u^2}\times\dfrac{du}{dx}\tag{20}\]
\[\dfrac{d}{dx}\cot^{-1}{u}=-\dfrac{1}{1+u^2}\times\dfrac{du}{dx}\tag{21}\]
\[\dfrac{d}{dx}\sec^{-1}{u}=\dfrac{1}{|u|\sqrt{u^2-1}}\times\dfrac{du}{dx}\qquad (|u|\gt 1)\tag{22}\]
\[\dfrac{d}{dx}\csc^{-1}{u}=-\dfrac{1}{|u|\sqrt{u^2-1}}\times\dfrac{du}{dx}\qquad (|u|\gt 1)\tag{23}\]
積分法 Integration
\[\int{kf(x)}\,dx=k\int{f(x)}\,dx\qquad (k\ne 0)\tag{1}\]
\[\int{f(x)+g(x)}\,dx=\int{f(x)}\,dx+\int{g(x)}\,dx\tag{2}\]
\[\int{u^n}\,dx=\dfrac{u^{n+1}}{n+1}+C\qquad (n\ne -1)\tag{3}\]
\[\int{\dfrac{du}{u}}=\ln{|u|}+C\tag{4}\]
\[\int{\cos{u}}\,du=\sin{u}+C\tag{5}\]
\[\int{\sin{u}}\,du=-\cos{u}+C\tag{6}\]
\[\int\tan{u}\,du=\ln{|\sec{u}|}+C=-\ln{|\cos{u}|}+C\tag{7}\]
\[\int\cot{u}\,du=\ln{|\sin{u}|}+C=-\ln{|\csc{u}|}+C\tag{8}\]
\[\int\sec^2{u}\,du=\tan{u}+C\tag{9}\]
\[\int\csc^2{u}\,du=-\cot{u}+C\tag{10}\]
\[\int\sec{u}\tan{u}\,du=\sec{u}+C\tag{11}\]
\[\int\csc{u}\cot{u}\,du=-\csc{u}+C\tag{12}\]
\[\int e^u \,du=e^u+C\tag{13}\]
\[\int a^u \,du=\dfrac{a^u}{\ln{a}}+C\qquad (a\gt 0, a\ne 1)\tag{14}\]
\[\int\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}{u}+C\tag{15}\]
\[\int\dfrac{du}{1+u^2}=\tan^{-1}{u}+C\tag{16}\]
\[\int\dfrac{du}{u\sqrt{u^2-1}}=\sec^{-1}{|u|}+C\tag{17}\]
65学院ではAP Calculus受験に向けた個別指導をおこなっています。詳しくはお問い合わせください。
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