数学の基礎と公式-4.2

数学の基礎と公式

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4. 微分-2

※$\log x$の$x$の箇所には絶対値がつきます。$\log | x |$

$\displaystyle{               }$

(1) $\displaystyle{\{\log(x^3+5x+2)\}'={{3\,x^2+5}\over{x^3+5\,x+2}}}$

(2) $\displaystyle{  (\log(x+\sqrt{x^2+1}))'={{1}\over{\sqrt{x^2+1}}}    }$

(3) $\displaystyle{ \left(\dfrac{1}{4} \log(\dfrac{x-2}{x+2}) \right)'    = {{1}\over{4\,\left(x-2\right)}}-{{1}\over{4\,\left(x+2\right)}}  }$

(4) $\displaystyle{  \left(\log \sqrt{\dfrac{1+x^2}{1-x^2}} \right)'      = -{{2\,x}\over{x^4-1}}    }$

(5) $\displaystyle{ (e^\frac{1}{x^2})'=-{{2\,e^{{{1}\over{x^2}}}}\over{x^3}}              }$

(6) $\displaystyle{ (x^2 e^x)'=x^2\,e^{x}+2\,x\,e^{x}              }$

(7) $\displaystyle{( x^2 e^\frac{1}{x})'=           2\,x\,e^{{{1}\over{x}}}-e^{{{1}\over{x}}}   }$

(8) $\displaystyle{  (4^{5x})'=5\,\log 4\,\times 4^{5\,x}             }$

(9) $\displaystyle{ (y^{\sqrt{x}})'={{y^{\sqrt{x}}\,\log y}\over{2\,\sqrt{x}}}              }$

(10) $\displaystyle{  (\sin ^{x}x)'=\sin ^{x}x\,\left(\log \sin x+{{x\,\cos x}\over{\sin x}}\right)             }$

(11) $\displaystyle{  (\cos ^{x}x)'=\cos ^{x}x\,\left(\log \cos x-{{x\,\sin x}\over{\cos x}}\right)            }$

(12) $\displaystyle{  (\tan ^{x}x)'=\tan ^{x}x\,\left(\log \tan x+{{x\,\left(\sec x\right)^2}\over{ \tan x}}\right)           }$

(13) $\displaystyle{  \left( \dfrac{\log(x)}{x^2}\right)'={{1}\over{x^3}}-{{2\,\log x}\over{x^3}}             }$

(14) $\displaystyle{   (\log(\sin x ))'={{\cos x}\over{\sin x}}            }$

(15) $\displaystyle{   (\log(\cos x ))'=-{{\sin x}\over{\cos x}}            }$

(16) $\displaystyle{   (\log(\tan x ))'={{\left(\sec x\right)^2}\over{\tan x}}            }$

(17) $\displaystyle{    (e^{-x} \sin x)'  =e^ {- x }\,\cos x-e^ {- x }\,\sin x         }$

(18) $\displaystyle{    (e^{-x} \cos x)'  =-e^ {- x }\,\sin x-e^ {- x }\,\cos x        }$

(19) $\displaystyle{    (e^{-x} \tan x)'  =e^ {- x }\,\left(\sec x\right)^2-e^ {- x }\,\tan x        }$

(20) $\displaystyle{    \left(\log \dfrac{\sqrt{x+5}+\sqrt{x+7}}{\sqrt{x+5}-\sqrt{x+7}} \right)'  ={{1}\over{\sqrt{x+5}\,\sqrt{x+7}}} }$

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