微分の公式

$\large1.\left( x^{p}\right) ‘=px^{p-1}$

$\large2.\left( \dfrac{1}{x}\right) ‘=-\dfrac{1}{x^{2}}$

$\large3.\left( e^{x}\right) ‘=e^{x}$

$\large4.\left( a^{x}\right) ‘=a^{x}\log a$

$\large5.\left( \log x\right) ‘=\dfrac{1}{x}$

$\large6.\left( \log _{a}x\right) ‘=\dfrac{1}{x\log a}$

$\large7.\left( \sin x\right) ‘=\cos x$

$\large8.\left( \cos x\right) ‘=-\sin x$

$\large9.\left( \tan x\right) ‘=\dfrac{1}{\cos ^{2}x}$

$\lim _\limits{h\rightarrow 0}\dfrac{\left( x+h\right) ^{p}-x^{p}}{h}$

二項定理より、
$\lim _\limits{h\rightarrow 0}\dfrac{x^{p}+_{p}\mathrm{C}_{1}x^{p-1}h+_{p}\mathrm{C}_{2}x^{p-2}h^{2}+\ldots +h^{p}-x^{p}}{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{_{p}\mathrm{C}_{1}x^{p-1}h+_{p}\mathrm{C}_{2}x^{p-2}h^{2}+\ldots +h^{p}}{h}$

$=\lim _\limits{h\rightarrow 0}{p}\mathrm{C}_{1}x^{p-1}+h\left( _{p}\mathrm{C}_{2}x^{p-2}+_{p}\mathrm{C}_{3}x^{p-3}h+\ldots +h^{p-2}\right) $

$=px^{p-1}$

$\lim _\limits{h\rightarrow 0}\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{-h}{\dfrac{x\left( x+h\right) }{h}}$

$=\lim _\limits{h\rightarrow 0}\dfrac{-1}{x\left( x+h\right) }$

$=-\dfrac{1}{x^{2}}$

$\lim _\limits{h\rightarrow 0}\dfrac{e^{x+h}-e^{x}}{h}$
$=\lim _\limits{h\rightarrow 0}e^{x}\times\left( \dfrac{e^{h}-1}{h}\right) $
$=e^{x}\cdot 1$
$=e^{x}$

$\lim _\limits{h\rightarrow 0}\dfrac{a^{x+h}-a^{x}}{h}$
$=\lim _\limits{h\rightarrow 0}a^{x}\times\left( \dfrac{a^{h}-1}{h}\right) $
$a^m=e^h$とおくと、$m=h\log a$となる。したがって、
$\lim _\limits{m\rightarrow 0}a^x\left( \dfrac{e^{m}-1}{\dfrac{m}{\log a}}\right) $
$=\lim _\limits{m\rightarrow 0}a^{x}\log ^{a}\left( \dfrac{e^{m}-1}{m}\right) $
$=a^{x}\log a\cdot 1$
$=a^{x}\log a$

$\lim _\limits{h\rightarrow 0}\dfrac{\log \left( x+h\right) -\log x}{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{1}{h}\log \left( \dfrac{x+h}{x}\right)$

$=\lim _\limits{h\rightarrow 0}\log \left( 1+\dfrac{h}{x}\right) ^{\dfrac{1}{h}} $

$=\lim _\limits{h\rightarrow 0}\dfrac{1}{x}\log \left( 1+\dfrac{h}{x}\right) ^{\dfrac{x}{h}}$

$=\dfrac{1}{x}\cdot 1$

$=\dfrac{1}{x}$

$\lim _\limits{h\rightarrow 0}\dfrac{\log_{a} \left( x+h\right) -\log_{a} x}{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{1}{h}\log_{a} \left( \dfrac{x+h}{x}\right)$

$=\lim _\limits{h\rightarrow 0}\log_{a} \left( 1+\dfrac{h}{x}\right) ^{\dfrac{1}{h}} $

$=\lim _\limits{h\rightarrow 0}\dfrac{1}{x}\log_{a} \left( 1+\dfrac{h}{x}\right) ^{\dfrac{x}{h}}$

対数の底の変換を行うと、

$=\lim _\limits{h\rightarrow 0}\dfrac{1}{x}\dfrac{log \left( 1+\dfrac{h}{x}\right) ^{\dfrac{x}{h}}}{log a}$

$=\dfrac{1}{x\cdot log a}\cdot 1$

$=\dfrac{1}{xlog a}$

$\lim _\limits{h\rightarrow 0}\dfrac{\sin \left( x+h\right) -\sin x}{h}$
三角関数の加法定理より、
$=\lim _\limits{h\rightarrow 0}\dfrac{\sin x\cosh +\cos x\sinh -\sin x}{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{\sin x\left( \cosh -1\right) }{h}+\cos x\dfrac{\sinh }{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{\sin x\left( \cos^{2}h -1\right) }{h(cosh+1)}+\cos x\dfrac{\sinh }{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{\sin x\sin^{2}h }{h(cosh+1)}+\cos x\dfrac{\sinh }{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{\sin x \sinh }{h}\cdot\dfrac{sinh}{(cosh+1)}+\cos x\cdot\dfrac{\sinh }{h}$

$=sinx\cdot 0 \cdot 1+cosx\cdot 1$

$=cosx$

$\lim _\limits{h\rightarrow 0}\dfrac{\cos \left( x+h\right) -\cos x}{h}$
三角関数の加法定理より、
$=\lim _\limits{h\rightarrow 0}\dfrac{\cos x\cosh -\sin x\sinh -\cos x}{h}$
$=\lim _\limits{h\rightarrow 0}\dfrac{\cos x\left( 1-\cosh \right) -\sin x\sinh }{h}$

7の微分より、
$=cosx\cdot 0\cdot 1 -sinx$
$=-sinx$

$\lim _\limits{h\rightarrow 0}\dfrac{\tan \left( x+h\right) -\tan x}{h}$
三角関数の加法定理より、
$=\lim _\limits{h\rightarrow 0}\dfrac{\dfrac{\tan x+\tanh }{1-\tan x\tanh }-\tan x}{h}$

$=\lim _\limits{h\rightarrow 0}\dfrac{\tan x+\tanh +\tan x^{2}\tanh -\tan x}{h\left( 1-\tan x\tanh \right) }$

$=\lim _\limits{h\rightarrow 0}\dfrac{\tanh \left( 1+\tan ^{2}x\right) }{h}\cdot \dfrac{1}{\left( 1-\tan x\tanh \right) }$

$=\lim _\limits{h\rightarrow 0}\dfrac{1}{\cos ^{2}x}\cdot \dfrac{\tanh }{h}\cdot \dfrac{1}{\left( 1-\tan x\tanh \right) }$

$=\dfrac{1}{\cos ^{2}x}$