1 \u5909\u6570 \\(x\\) \u306e\u95a2\u6570 \\(y = f(x) \\) \u306e\u300c\u5c0e\u95a2\u6570<\/strong><\/span>\u300d\u3068\u306f<\/p>\n $$ \\frac{df}{dx} \\equiv \\lim_{h \\rightarrow 0} \\frac{f(x + h) -f(x)} {(x + h) -x} = \\lim_{h \\rightarrow 0} \\frac{f(x + h) -f(x)} { h}$$<\/p>\n \u3064\u307e\u308a\uff0c\u5909\u6570 \\(x\\) \u304c \\(x\\) \u304b\u3089 \\(x + h\\) \u306b\u5909\u5316\u3059\u308b\u3068\u304d\uff0c\u95a2\u6570 \\( f(x)\\) \u306e\u5909\u5316\u5206 \\( f(x + h) -f(x) \\) \u3092\u5909\u6570 \\( x\\) \u306e\u5909\u5316\u5206 \\( (x+h) -x = h\\) \u3067\u5272\u3063\u305f\u5909\u5316\u7387\u306e \\(h \\rightarrow 0\\) \u306e\u6975\u9650\u3002<\/p>\n \\(f(x)\\) \u306e\u5c0e\u95a2\u6570\u3092\u6c42\u3081\u308b\u3053\u3068\u3092\u300c\\(f(x)\\) \u3092\u5fae\u5206\u3059\u308b\u300d\u3068\u3044\u3046\u3002<\/p>\n \\( y = f(x) \\) \u306e\u5c0e\u95a2\u6570\u306e\u8868\u8a18\u4f8b\uff1a<\/p>\n $$ \\frac{dy}{dx}, \\quad y’, \\quad \\frac{df}{dx}, \\quad f'(x), \\quad \\left\\{f(x)\\right\\}’, \\quad \\frac{d}{dx} f(x), \\mbox{etc.} \\ \\ \\dots$$<\/p>\n \u6ce8\u610f\uff1a<\/p>\n \u529b\u5b66\u7b49\u3067 \\(x(t), y(t)\\) \u306e\u3088\u3046\u306b\u5909\u6570 \\(t\\) \u306e\u95a2\u6570\u3067\u3042\u308b\u5834\u5408\uff0c<\/p>\n $$ \\frac{dx}{dt} = \\dot{x}, \\ \\ \\frac{dy}{dt} = \\dot{y} $$ \u306e\u3088\u3046\u306b\uff0c\\(\\dot{\\ } \\) \uff08\u30c9\u30c3\u30c8\uff09\u3067\u5fae\u5206\u3092\u8868\u3059\u3002\u3082\u3057\uff0c\\( {\\ }’ \\) \uff08\u30c0\u30c3\u30b7\u30e5\uff0c\u30d7\u30e9\u30a4\u30e0\uff09\u3067\u5fae\u5206\u3092\u8868\u3059\u3068<\/p>\n $$ x’ = \\frac{dx}{dx} = 1$$ \u3068\u52d8\u9055\u3044\u3055\u308c\u308b\u304b\u3089\u3002<\/p>\n \u9ad8\u6728\u8c9e\u6cbb\u8457\u300c\u89e3\u6790\u6982\u8ad6\u300d\u306b\u3088\u308c\u3070\uff0c\u8a18\u53f7 \\(\\displaystyle \\frac{dy}{dx}\\) \u306f\u30e9\u30a4\u30d7\u30cb\u30c3\u30c4\u304b\u3089\u306e\u4f1d\u7d71\uff0c\\(f'(x) \\) \u306f\u30e9\u30b0\u30e9\u30f3\u30b8\u30e5\u306e\u7528\u4f8b\uff0c\\( \\dot{x}, \\dot{y} \\) \u306f\u30cb\u30e5\u30fc\u30c8\u30f3\u306e\u7528\u4f8b\u3067\u3042\u308b\u3068\u8a18\u8f09\u304c\u3042\u308b\u3002<\/p>\n \\( f(x) = x^2 \\) \u3092\u5fae\u5206\u3059\u308b\u3002<\/p>\n \\begin{eqnarray} \u5b9a\u7fa9 <\/p>\n 1 \u5909\u6570 \\(x\\) \u306e\u95a2\u6570 \\(y = f(x) \\) \u306e\u300c\u5c0e\u95a2\u6570\u300d\u3068\u306f<\/p>\n $$ \\frac{df}{dx} \\equiv \\lim_{h \\rightarrow 0} \\frac{f(x + h) -f(x)} {(x + h) -x} = \\lim_{h \\rightarrow 0} \\frac{f(x + h) -f(x)} { h}$$<\/p>\n \u3064\u307e\u308a\uff0c\u5909\u6570 \\(x\\) \u304c \\(x\\) \u304b\u3089 \\(x + h\\) \u306b\u5909\u5316\u3059\u308b\u3068\u304d\uff0c\u95a2\u6570 \\( f(x)\\) \u306e\u5909\u5316\u5206 \\( f(x + h) -f(x) \\) \u3092\u5909\u6570 \\( x\\) \u306e\u5909\u5316\u5206 \\( (x+h) -x = h\\) \u3067\u5272\u3063\u305f\u5909\u5316\u7387\u306e \\(h \\rightarrow 0\\) \u306e\u6975\u9650\u3002<\/p>\n","protected":false},"author":33,"featured_media":0,"parent":2068,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"inline_featured_image":false,"footnotes":""},"class_list":["post-2073","page","type-page","status-publish","hentry","nodate","item-wrap"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\u8868\u8a18<\/h3>\n
\u4f8b<\/h3>\n
\n\\frac{df}{dx} &=& \\lim_{h \\rightarrow 0} \\frac{(x + h)^2 -x^2}{h} = \\lim_{h \\rightarrow 0} \\frac{(x^2 + 2 h x + h^2) -x^2}{h} \\\\
\n&=& \\lim_{h \\rightarrow 0} \\frac{2 h x + h^2}{h} \\\\
\n&=& \\lim_{h \\rightarrow 0} (2 x + h) = 2 x
\n\\end{eqnarray}<\/p>\n","protected":false},"excerpt":{"rendered":"