微積分の問題の解答

微分の問題($a,c,n$は定数)

$(c)^\prime=$$0$ $\qquad (x^n)^\prime=$$nx^{n-1}$
$(e^x)^\prime=$$e^x$ $\qquad (e^{ax})^\prime=$$ae^{ax}$ $\qquad (\log x)^\prime=$$\dfrac{1}{x}$ $\qquad$
$(\sin x)^\prime=$$\cos x$ $(\cos x)^\prime=$$-\sin x$ $(\sin ax)^\prime=$$a\cos ax$ $(\cos ax)^\prime=$$-a\sin ax$
逆三角関数の微分: $(\sin^{-1} x)^\prime=$$\dfrac{1}{\sqrt{1-x^2}}$ $(\cos^{-1} x)^\prime=$$-\dfrac{1}{\sqrt{1-x^2}}$ $(\tan^{-1}x)^\prime=$$\dfrac{1}{1+x^2}$ $\quad$
合成関数の微分: $y=f(g(x)), u=g(x)$とおくと$y=f(u)$で, このとき,
$\dfrac{dy}{dx}=$$\dfrac{dy}{du}$$\times$$\dfrac{du}{dx}$, 又は, $y^\prime=$$f^\prime(u)$$\times$$u^\prime$
積・商の微分: $(f\cdot g)^\prime=$$f^\prime\cdot g+f\cdot g^\prime$ $ \left(\dfrac{f}{g}\right)^\prime=$$\dfrac{f^\prime\cdot g-f\cdot g^\prime}{g^2}$

積分の問題($a,b$は定数, $C$は積分定数)

$a\ne -1$のとき, $\displaystyle\int x^adx=$$\dfrac{1}{a+1}x^{a+1}+C$
$\displaystyle\int \dfrac{1}{x}dx=$$\log|x|+C$ $\displaystyle\int \dfrac{1}{x+a}dx=$$\log|x+a|+C$ $\displaystyle\int \dfrac{f^\prime(x)}{f(x)}dx=$$\log|f(x)|+C$
$\displaystyle\int \dfrac{1}{1+x^2}dx=$$\tan^{-1}x+C$ $\displaystyle\int \dfrac{x}{1+x^2}dx=$$\dfrac{1}{2}\log(1+x^2)+C$
$\displaystyle\int e^xdx=$$e^x+C$ $\displaystyle\int e^{ax}dx=$$\dfrac{1}{a}e^{ax}+C$
$\displaystyle\int \sin xdx=$$-\cos x+C$ $\displaystyle\int \cos xdx=$$\sin x+C$ $\displaystyle\int \sin axdx=$$-\dfrac{1}{a}\cos ax+C$ $\displaystyle\int \cos axdx=$$\dfrac{1}{a}\sin ax+C$
置換積分: $x=g(u)$とおくと$\displaystyle\int f(x)dx=\int$$f(g(u))g^\prime(u)$$du$
部分積分: $\displaystyle\int f\cdot g^\prime dx=$$\displaystyle f\cdot g-\int f^\prime\cdot gdx$
$\displaystyle\int e^{ax}\sin bxdx=$$\dfrac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C$ $\displaystyle\int e^{ax}\cos bxdx=$$\dfrac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+C$