{"id":1940,"date":"2019-03-16T11:57:10","date_gmt":"2019-03-16T02:57:10","guid":{"rendered":"https:\/\/sasamath.com\/wp\/?p=1940"},"modified":"2019-09-05T19:49:28","modified_gmt":"2019-09-05T10:49:28","slug":"calculus-differentiation-rules","status":"publish","type":"post","link":"https:\/\/sasamath.com\/blog\/articles\/calculus-differentiation-rules\/","title":{"rendered":"\uc0ac\uce59\uacc4\uc0b0\uacfc \uad00\ub828\ub41c \ubbf8\ubd84 \ubc95\uce59"},"content":{"rendered":"

\uc774 \ud3ec\uc2a4\ud2b8\uc5d0\uc11c\ub294 \uc0ac\uce59\uacc4\uc0b0\uacfc \uad00\ub828\ub41c \ubbf8\ubd84\uc758 \uacc4\uc0b0 \ubc95\uce59\uc744 \uc720\ub3c4\ud558\uace0 \ub2e4\ud56d\ud568\uc218\uc640 \uc720\ub9ac\ud568\uc218\uc758 \ubbf8\ubd84\ubc95\uc744 \uc0b4\ud3b4\ubcf8\ub2e4.<\/p>\n

\uba3c\uc800 \uc0c1\uc218\ud568\uc218\uc640 \ub2e8\ud56d\ud568\uc218\uc758 \ubbf8\ubd84\ubc95\uc744 \uc0b4\ud3b4\ubcf4\uc790.<\/p>\n

\n

\uc815\ub9ac 1. (\uc0c1\uc218\ud568\uc218\uc758 \ubbf8\ubd84)<\/span><\/p>\n

\\(f\\)\uac00 \uc0c1\uc218\ud568\uc218\uc774\uace0 \\(f(x)=c\\)\uc77c \ub54c \\(f ‘ (x) = 0\\)\uc774\ub2e4.<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\\[f ‘ (x) = \\lim_{h\\to 0} \\frac{f(x+h)-f(x)}{h} = \\lim_{h\\to 0}\\frac{c-c}{h} = 0.\\tag*{\\(\\blacksquare\\)}\\]\n<\/p>\n<\/div>\n

\n

\uc815\ub9ac 2. (\uac70\ub4ed\uc81c\uacf1 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\\(n\\)\uc774 \uc790\uc5f0\uc218\uc774\uace0 \\(f(x)=x^n\\)\uc77c \ub54c \\(f ‘ (x) = nx^{n-1}\\)\uc774\ub2e4. (\ub2e8, \\(n=1,\\) \\(x=0\\)\uc77c \ub54c\uc5d0\ub294 \\(f ‘ (0) = 1\\)\uc774\ub2e4.)<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\uba3c\uc800 \\(n \\ge 2\\)\uc778 \uacbd\uc6b0\ub97c \uc99d\uba85\ud558\uc790. \\(t\\ne x\\)\uc77c \ub54c
\n\\[t^n – x^n = (t-x)(t^{n-1} + t^{n-2}x + \\cdots + tx^{n-2} + x^{n-1})\\]
\n\uc774\ubbc0\ub85c \ub2e4\uc74c\uc744 \uc5bb\ub294\ub2e4.
\n\\[\\begin{align}
\nf ‘ (x) &= \\lim_{t\\to x}\\frac{f(t)-f(x)}{t-x} \\\\[6pt]
\n&= \\lim_{t\\to x}\\frac{t^n – x^n}{t-x} \\\\[6pt]
\n&= \\lim_{t\\to x}\\frac{(t-x)(t^{n-1} + t^{n-2}x + \\cdots + tx^{n-2} + x^{n-1})}{t-x} \\\\[6pt]
\n&= \\lim_{t\\to x}(t^{n-1} + t^{n-2}x + \\cdots + tx^{n-2} + x^{n-1}) \\\\[6pt]
\n&= x^{n-1} + x^{n-2}x + \\cdots + x^{n-1} \\quad (n \\text{ terms})\\\\[6pt]
\n&= nx^{n-1}.
\n\\end{align}\\]
\n\\(n=1\\)\uc77c \ub54c\uc5d0\ub294
\n\\[f ‘ (x) = \\lim_{t\\to x}\\frac{f(t)-f(x)}{t-x} = \\lim_{t\\to x}\\frac{t-x}{t-x} = 1\\]
\n\uc774\ubbc0\ub85c \\(f ‘ (x) = 1\\)\uc774\ub2e4. \ud2b9\ud788 \\(x\\ne 0\\)\uc77c \ub54c \\(f ‘ (x) = 1 = x^0\\)\uc774\ub2e4.
\n<\/span><\/p>\n<\/div>\n

\n

\ucc38\uace0.<\/span>
\n\uc0ac\uc2e4 \uc815\ub9ac 2\uc758 \uacf5\uc2dd\uc740 \uc9c0\uc218 \\(n\\)\uc774 \uc2e4\uc218\uc77c \ub54c\uc5d0\ub3c4 \uc0ac\uc6a9\ud560 \uc218 \uc788\ub2e4. \uc774 \ud3ec\uc2a4\ud2b8\uc5d0\uc11c\ub294 \uc9c0\uc218\uac00 \uc815\uc218\uc77c \ub54c\uc758 \uacf5\uc2dd\uc744 \uc99d\uba85\ud558\uba70(\ub530\ub984\uc815\ub9ac 4), \uc9c0\uc218\uac00 \uc2e4\uc218\uc77c \ub54c\uc758 \uacf5\uc2dd\uc740 \ub85c\uadf8\ud568\uc218\uc758 \ubbf8\ubd84\ubc95\uc744 \uc0b4\ud3b4\ubcf8 \ub4a4 \uc99d\uba85\ud560 \uac83\uc774\ub2e4.
\n<\/span><\/p>\n<\/div>\n

\n

\uc815\ub9ac 3. (\uc0c1\uc218\ubc30 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\\(c\\)\uac00 \uc0c1\uc218\ud568\uc218\uc774\uace0 \\(f\\)\uac00 \ubbf8\ubd84 \uac00\ub2a5\ud55c \ud568\uc218\uc77c \ub54c \ud568\uc218 \\(cf\\)\uc758 \ub3c4\ud568\uc218\ub294
\n\\[\\frac{d}{dx}(cf) = c\\frac{df}{dx} .\\]\n<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\\[\\begin{align}
\n\\frac{d}{dx}(cf)
\n&= \\lim_{h\\to 0}\\frac{cf(x+h) – cf(x)}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{c(f(x+h) – f(x))}{h} \\\\[6pt]
\n&= c\\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h} \\\\[6pt]
\n&= c \\frac{df}{dx}.\\tag*{\\(\\blacksquare\\)}
\n\\end{align}\\]<\/p>\n<\/div>\n

\uc815\ub9ac 2\uc640 \uc815\ub9ac 3\uc744 \uacb0\ud569\ud558\uba74 \ub2e4\uc74c\uc744 \uc5bb\ub294\ub2e4.<\/p>\n

\n

\ub530\ub984\uc815\ub9ac 1. (\ub2e8\ud56d\ud568\uc218\uc758 \ubbf8\ubd84)<\/span><\/p>\n

\\(c\\)\uac00 \uc0c1\uc218\uc774\uace0 \\(n\\)\uc774 \uc790\uc5f0\uc218\uc77c \ub54c
\n\\[\\frac{d}{dx}(cx^n ) = cnx^{n-1}.\\]
\n\ub2e8, \ud3b8\uc758\uc0c1 \\(n=1,\\) \\(x=0\\)\uc77c \ub54c \uc6b0\ubcc0\uc740 \\(0^0 = 1\\)\ub85c \uacc4\uc0b0\ud55c\ub2e4.<\/p>\n<\/div>\n

\ub2e4\uc74c\uc73c\ub85c \uc0ac\uce59\uacc4\uc0b0\uacfc \uad00\ub828\ub41c \ubbf8\ubd84\uc758 \uacc4\uc0b0 \ubc95\uce59\uc744 \uc0b4\ud3b4\ubcf4\uc790.<\/p>\n

\n

\uc815\ub9ac 4. (\ub367\uc148 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\ub450 \ud568\uc218 \\(f\\)\uc640 \\(g\\)\uac00 \ubbf8\ubd84 \uac00\ub2a5\ud558\uace0 \\(y=f(x)+g(x)\\)\uc77c \ub54c
\n\\[y ‘ = f ‘ (x) + g ‘ (x) .\\]\n<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\\[\\begin{align}
\ny ‘
\n&= \\lim_{h\\to 0}\\frac{(f(x+h)+g(x+h)) – (f(x)+g(x))}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{(f(x+h)-f(x)) + (g(x+h)-g(x))}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h} + \\lim_{h\\to 0}\\frac{g(x+h)-g(x)}{h} \\\\[6pt]
\n&= f ‘ (x) + g ‘ (x) . \\tag*{\\(\\blacksquare\\)}
\n\\end{align}\\]\n<\/p>\n<\/div>\n

\n

\ub530\ub984\uc815\ub9ac 2. (\ube84\uc148 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\ub450 \ud568\uc218 \\(f\\)\uc640 \\(g\\)\uac00 \ubbf8\ubd84 \uac00\ub2a5\ud558\uace0 \\(y=f(x)-g(x)\\)\uc77c \ub54c
\n\\[y ‘ = f ‘ (x) – g ‘ (x) .\\]\n<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\\(h(x) = -g(x)\\)\ub77c\uace0 \ud558\uba74 \\(y = f(x)+h(x)\\)\uc774\uace0, \uc815\ub9ac 3\uc5d0 \uc758\ud558\uc5ec \\(h ‘ (x) = – g ‘ (x)\\)\uc774\ubbc0\ub85c
\n\\[y ‘ = f ‘ (x) + h ‘ (x) = f ‘ (x) – g ‘ (x)\\]
\n\uc774\ub2e4.<\/span><\/p>\n<\/div>\n

\n

\ub530\ub984\uc815\ub9ac 3. (\ub2e4\ud56d\ud568\uc218\uc758 \ubbf8\ubd84)<\/span><\/p>\n

\\(f\\)\uac00 \ub2e4\ud56d\ud568\uc218\uc774\uace0 \uc0c1\uc218 \\(a_n ,\\) \\(a_{n-1} ,\\) \\(\\cdots ,\\) \\(a_0\\)\uc5d0 \ub300\ud558\uc5ec
\n\\[f(x)=a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_1 x + a_0\\]
\n\uaf34\ub85c \ub098\ud0c0\ub098\uba74
\n\\[f ‘ (x) = na_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \\cdots + a_1 \\tag{1}\\]
\n\uc774\ub2e4.\n<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\\(n\\)\uc5d0 \ub300\ud55c \uc218\ud559\uc801 \uadc0\ub0a9\ubc95\uc744 \uc0ac\uc6a9\ud558\uc790. \\(n=1\\)\uc77c \ub54c\uc5d0\ub294
\n\\[f(x) = a_1 x + a_0\\]
\n\uc774\ubbc0\ub85c
\n\\[f ‘ (x) = a_1\\]
\n\ub85c\uc11c (1)\uc774 \uc131\ub9bd\ud55c\ub2e4.<\/p>\n

\uc774\uc81c \\(k\\)\uac00 \uc790\uc5f0\uc218\uc774\uace0 \\(n=k\\)\uc77c \ub54c (1)\uc774 \uc131\ub9bd\ud55c\ub2e4\uace0 \uac00\uc815\ud558\uc790. \uadf8\ub9ac\uace0
\n\\[f(x)=a_{k+1} x^{k+1} + a_k x^k + a_{k-1} x^{k-1} + \\cdots + a_1 x + a_0\\]
\n\uc774\ub77c\uace0 \ud558\uc790. \uadf8\ub7ec\uba74
\n\uc815\ub9ac 4\uc640 \\(n=k\\)\uc77c \ub54c\uc758 \uadc0\ub0a9\uc801 \uac00\uc815\uc5d0 \uc758\ud558\uc5ec
\n\\[\\begin{align}
\nf ‘ (x)
\n&= \\frac{d}{dx} (a_{k+1} x^{k+1}) + \\frac{d}{dx}(a_n x^k + a_{k-1} x^{k-1} + \\cdots + a_1 x + a_0)\\\\[6pt]
\n&= (k+1)a_{k+1} x^k + (ka_k x^{k-1} + (k-1)a_{k-1}x^{k-2} + \\cdots + a_1) \\\\[8pt]
\n&= (k+1)a_{k+1} x^k + ka_k x^{k-1} + (k-1)a_{k-1}x^{k-2} + \\cdots + a_1
\n\\end{align}\\]
\n\uc774\ubbc0\ub85c \\(n=k+1\\)\uc77c \ub54c\uc5d0\ub3c4 (1)\uc774 \uc131\ub9bd\ud55c\ub2e4.<\/p>\n

\uadf8\ub7ec\ubbc0\ub85c \uc218\ud559\uc801 \uadc0\ub0a9\ubc95\uc5d0 \uc758\ud558\uc5ec \uc784\uc758\uc758 \uc790\uc5f0\uc218 \\(n\\)\uc5d0 \ub300\ud558\uc5ec (1)\uc774 \uc131\ub9bd\ud55c\ub2e4.
\n<\/span><\/p>\n<\/div>\n

\n

\ubcf4\uae30 1.<\/span>
\n\ub2e4\uc74c \ud568\uc218\uc758 \ub3c4\ud568\uc218\ub97c \uad6c\ud574\ubcf4\uc790.
\n\\[\\begin{align}
\nf(x) &= 3x^2 -4x +7, \\\\[8pt]
\ng(x) &= -5x^7 +3x^4 -4x +1.
\n\\end{align}\\]
\n\ub530\ub984\uc815\ub9ac 3\uc744 \uc774\uc6a9\ud558\uba74 \ub2e4\uc74c\uacfc \uac19\uc774 \uacc4\uc0b0\ub41c\ub2e4.
\n\\[\\begin{align}
\nf ‘ (x) &= 6x -4, \\\\[8pt]
\ng ‘ (x) &= -35x^6 +12x^3 -4. \\tag*{\\(\\square\\)}
\n\\end{align}\\]\n<\/p>\n<\/div>\n

\n

\uc815\ub9ac 5. (\uacf1\uc148 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\ub450 \ud568\uc218 \\(f\\)\uc640 \\(g\\)\uac00 \ubbf8\ubd84 \uac00\ub2a5\ud558\uace0 \\(y=f(x)g(x)\\)\uc77c \ub54c
\n\\[y ‘ = f ‘ (x) g (x) + f(x) g ‘ (x) .\\]\n<\/p><\/div>\n

\n

\uc99d\uba85<\/p>\n

\\[\\begin{align}
\ny ‘
\n&= \\lim_{h\\to 0}\\frac{f(x+h)g(x+h) – f(x)g(x)}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{f(x+h)g(x+h) – f(x)g(x+h) + f(x)g(x+h) – f(x)g(x)}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{(f(x+h)-f(x))g(x+h) + f(x)(g(x+h) – g(x))}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{f(x+h)-f(x)}{h} \\cdot \\lim_{h\\to 0} g(x+h) + f(x) \\cdot \\lim_{h\\to 0}\\frac{g(x+h) – g(x)}{h} \\\\[6pt]
\n&= f ‘ (x) g(x) + f(x)g ‘ (x) . \\tag*{\\(\\blacksquare\\)}
\n\\end{align}\\]\n<\/p>\n<\/div>\n

\n

\uc815\ub9ac 6. (\ub098\ub217\uc148 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\ub450 \ud568\uc218 \\(f\\)\uc640 \\(g\\)\uac00 \ubbf8\ubd84 \uac00\ub2a5\ud558\uace0
\n\\[y=\\frac{f(x)}{g(x)}\\]
\n\ub77c\uace0 \ud558\uc790. \uc774\ub54c \\(g (x) \\ne 0\\)\uc778 \\(x\\)\uc5d0 \ub300\ud558\uc5ec
\n\\[y ‘ = \\frac{f ‘ (x) g (x) – f(x) g ‘ (x)}{(g(x))^2} .\\]\n<\/p><\/div>\n

\n

\uc99d\uba85<\/p>\n

\\(h(x) = 1\/g(x)\\)\uc774\ub77c\uace0 \ud558\uba74
\n\\[\\begin{align}
\nh ‘ (x)
\n&= \\lim_{h\\to 0}\\frac{h(x+h) – h(x)}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{1\/g(x+h) -1\/g(x)}{h} \\\\[6pt]
\n&= \\lim_{h\\to 0}\\frac{g(x) – g(x+h)}{hg(x+h)g(x)} \\\\[6pt]
\n&= -\\lim_{h\\to 0}\\frac{g(x+h)-g(x)}{h} \\cdot \\lim_{h\\to 0}\\frac{1}{g(x+h)g(x)} \\\\[6pt]
\n&= – g ‘ (x) \\cdot \\frac{1}{(g(x)^2} \\\\[6pt]
\n&= – \\frac{g ‘ (x)}{(g(x))^2}
\n\\end{align}\\]
\n\uc774\uace0, \\(y=f(x)h(x)\\)\uc774\ubbc0\ub85c
\n\\[\\begin{align}
\ny ‘
\n&= f ‘ (x) h(x) + f(x) h ‘ (x) \\\\[6pt]
\n&= \\frac{f ‘ (x)}{g(x)} – \\frac{f(x) g ‘ (x)}{(g(x))^2} \\\\[6pt]
\n&= \\frac{f ‘ (x)g(x)}{(g(x))^2} – \\frac{f(x) g ‘ (x)}{(g(x))^2} \\\\[6pt]
\n&= \\frac{f ‘ (x) g(x) – f(x) g ‘ (x)}{(g(x))^2}
\n\\end{align}\\]
\n\uc774\ub2e4.<\/span><\/p>\n<\/div>\n

\n

\ub530\ub984\uc815\ub9ac 4. (\uc74c\uc758 \uc815\uc218 \uc9c0\uc218\uc5d0 \ub300\ud55c \uac70\ub4ed\uc81c\uacf1 \ubbf8\ubd84 \ubc95\uce59)<\/span><\/p>\n

\\(m\\)\uc774 \uc74c\uc758 \uc815\uc218\uc774\uace0 \\(f(x)=x^m\\)\uc77c \ub54c \\(f ‘ (x) = mx^{m-1}\\)\uc774\ub2e4.<\/p>\n<\/div>\n

\n

\uc99d\uba85<\/p>\n

\\(n=-m,\\) \\(g(x) = x^{n}\\)\uc774\ub77c\uace0 \ud558\uba74 \uc815\ub9ac 2\uc5d0 \uc758\ud558\uc5ec
\n\\[g ‘ (x) = nx^{n-1}\\]
\n\uc774\ub2e4. \ub610\ud55c \\(f(x) = 1\/g(x)\\)\uc774\ubbc0\ub85c \uc815\ub9ac 6\uc5d0 \uc758\ud558\uc5ec
\n\\[f ‘ (x) = -\\frac{g ‘ (x)}{(g(x))^2} = -\\frac{nx^{n-1}}{x^{2n}} = -nx^{-n-1} = mx^{m-1}\\]
\n\uc774\ub2e4.
\n<\/span><\/p>\n<\/div>\n

\uc720\ub9ac\uc2dd\uc740 \ubb38\uc790, \uc0c1\uc218, \uc0ac\uce59\uacc4\uc0b0\uc73c\ub85c \uc774\ub8e8\uc5b4\uc9c4 \uc2dd\uc774\ubbc0\ub85c \uc9c0\uae08\uae4c\uc9c0 \uc0b4\ud3b4\ubcf8 \ubbf8\ubd84 \ubc95\uce59\uc744 \uc774\uc6a9\ud558\uba74 \ubcc0\uc218\uac00 \ud558\ub098\uc778 \ubaa8\ub4e0 \uc720\ub9ac\ud568\uc218\uc758 \ub3c4\ud568\uc218\ub97c \uad6c\ud560 \uc218 \uc788\ub2e4.<\/p>\n

\n

\ubcf4\uae30 2.<\/span>
\n\ub2e4\uc74c \ud568\uc218\uc758 \ub3c4\ud568\uc218\ub97c \uad6c\ud574\ubcf4\uc790.
\n\\[f(x) = \\frac{x^2 -x+3}{2x-5} \\quad \\left(x\\ne \\frac{5}{2}\\right)\\]
\n\ub098\ub217\uc148 \ubbf8\ubd84 \ubc95\uce59\uacfc \ub2e4\ud56d\ud568\uc218\uc758 \ubbf8\ubd84 \uacf5\uc2dd\uc744 \uc774\uc6a9\ud558\uba74 \ub2e4\uc74c\uc744 \uc5bb\ub294\ub2e4.
\n\\[\\begin{align}
\nf ‘ (x)
\n&= \\frac{(2x-1)(2x-5) – (x^2-x+3)\\times 2}{ (2x-5)^2} \\\\[6pt]
\n&= \\frac{4x^2 -12x +5 -2x^2 +2x -6}{(2x-5)^2} \\\\[6pt]
\n&= \\frac{2x^2 -10x -1}{4x^2 -20x +25} .\\tag*{\\(\\square\\)}
\n\\end{align}\\]\n<\/p>\n<\/div>\n

\ubd84\ubaa8\uac00 \ub2e8\ud56d\uc2dd\uc77c \ub550 \ubcf4\uae30 2\uc640\ub294 \ub2e4\ub978 \ubc29\ubc95\uc73c\ub85c \ub3c4\ud568\uc218\ub97c \uad6c\ud560 \uc218 \uc788\ub2e4.<\/p>\n

\n

\ubcf4\uae30 3.<\/span>
\n\ub2e4\uc74c \ud568\uc218\uc758 \ub3c4\ud568\uc218\ub97c \uad6c\ud574\ubcf4\uc790.
\n\\[g(x) = \\frac{x^2 -x+3}{2x} \\quad (x \\ne 0)\\]
\n\uc2dd\uc744 \ubcc0\ud615\ud558\uba74
\n\\[g(x) = \\frac{x}{2} -2 + \\frac{3}{2x}\\]
\n\uc774\ubbc0\ub85c
\n\\[g ‘ (x) = \\frac{1}{2} – \\frac{3}{2x^2} = \\frac{x^2 -3}{2x^2}. \\tag*{\\(\\square\\)}\\]\n<\/p>\n<\/div>\n

<\/p>\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

\uc774 \ud3ec\uc2a4\ud2b8\uc5d0\uc11c\ub294 \uc0ac\uce59\uacc4\uc0b0\uacfc \uad00\ub828\ub41c \ubbf8\ubd84\uc758 \uacc4\uc0b0 \ubc95\uce59\uc744 \uc720\ub3c4\ud558\uace0 \ub2e4\ud56d\ud568\uc218\uc640 \uc720\ub9ac\ud568\uc218\uc758 \ubbf8\ubd84\ubc95\uc744 \uc0b4\ud3b4\ubcf8\ub2e4. \uba3c\uc800 \uc0c1\uc218\ud568\uc218\uc640 \ub2e8\ud56d\ud568\uc218\uc758 \ubbf8\ubd84\ubc95\uc744 \uc0b4\ud3b4\ubcf4\uc790. \uc815\ub9ac 1. (\uc0c1\uc218\ud568\uc218\uc758 \ubbf8\ubd84) \\(f\\)\uac00 \uc0c1\uc218\ud568\uc218\uc774\uace0 \\(f(x)=c\\)\uc77c \ub54c \\(f ‘ (x) = 0\\)\uc774\ub2e4. \uc99d\uba85 \\(f ‘ (x) = \\lim_{h\\to 0} \\frac{f(x+h)-f(x)}{h} = \\lim_{h\\to 0}\\frac{c-c}{h} = 0.\\tag*{\\(\\blacksquare\\)}\\) \uc815\ub9ac 2. (\uac70\ub4ed\uc81c\uacf1 \ubbf8\ubd84 \ubc95\uce59) \\(n\\)\uc774 \uc790\uc5f0\uc218\uc774\uace0 \\(f(x)=x^n\\)\uc77c \ub54c \\(f ‘ (x) = nx^{n-1}\\)\uc774\ub2e4. (\ub2e8, \\(n=1,\\) \\(x=0\\)\uc77c \ub54c\uc5d0\ub294 \\(f ‘ (0)…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","footnotes":""},"categories":[47],"tags":[189,199,198],"class_list":["post-1940","post","type-post","status-publish","format-standard","hentry","category-calculus-ap","tag-differentiation","tag-differentiation-rule","tag-198"],"_links":{"self":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/1940","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/comments?post=1940"}],"version-history":[{"count":41,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/1940\/revisions"}],"predecessor-version":[{"id":2830,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/1940\/revisions\/2830"}],"wp:attachment":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/media?parent=1940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/categories?post=1940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/tags?post=1940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}