対数関数の導関数:

       ・(log x)’=1/x        (log a x)’=1/(xloga)       {log|f(x)|}’=f ’(x)/f(x)

   ・{logc(ax+b)}’=(ax+b)’/( ax+b)logc     (cf){(ax+b)’(ax+b)}dx=log|ax+b|+C

lim(h0)(1+h)1/h=e (e2.71828…)…これはlim(h→∞){1+(1/x)}x=eとも表せる。       

対数関数y=logx, y=logaxの導関数:    (log x)’=1/x,  (logax)’=1/(xloga)

    ○(log x)’=1/x,(x>0)の証明: 

         ☞f(x)=logxとおいて定義にしたがって微分すると

    f ’(x)= lim(h0){f(x+h)f(x)}/h= lim(h0){log(x+h)logx}/h

         ここで対数法則を使うと 

        与式= lim(h0)[log{(x+h)/x}/h= lim(h0) (1/h)log{1+(h/x)}

   = lim(h0)log{1+(h/x)}1/h                                                   

極限公式lim(x0)( ex1)/x=1, lim(x0)(1+x)1/x=eの後者を用いるために

さらに変形すると与式= lim(h0)log[{1+(h/x)}x/h]1/x= lim(h0)(1/x)log[{1+(h/x)}x/h]

ここで極限公式を用いると与式= lim(h0) (1/x)loge= lim(h0)(1/x)1=1/x//

    ○(log |x|)’=1/xの証明:                                                            

       ☞( i) x>0のとき、y=logxであるからy’=1/x  

   (ii) x<0のとき、y=log(x) であるからy’=1/(x)(x)’=1/x

        よって(log |x|)’=1/x//

〔別解〕( i)x>0のときは (log x)’=1/x,(x>0)の証明と同じ。    

              (ii) x<0としてx=uとおくとu>0 よってy=log|x|=log|u |+logu

           ここで合成関数の微分法y’=f ’(u)uを用いて微分すると

           y’=(logu)’(x)’  u>0より(logu)’=1/uを使うと

           y’=(1/u)(1)=1/(x)(1)=1/x//                          

   ○(logax)’=1/(xloga)の証明: 

  ☞(logax)’=lim(x0)[{loga(x+x)logax}/x]

               = lim(x0)(1/x)loga{1+(x/x)} 

  ここで⊿x/x =hとおくと⊿x0のときh0であるから

     (logax)’=lim(h0)[{1/(xh)}loga(1+h)]=(1/x)lim(h0)loga(1+h)1/h

               =(1/x)logae=1/(xlogea)=1/(xloga)//

〔別解〕☞y’=1/(loga|x|)(1/x)=1/(xloga)//

(log x)’=1/x//               {logc(ax+b)}’=(ax+b)’/( ax+b)logc

{log(ax+b)}’=1/(ax+b)(ax+b)’=(ax+b)’/(ax+     (cf)logea=b」⇆「eb=a

{log10(ax+b)}’=[1/{(ax+b)log10}](ax+b)’=(ax+b)’/{(ax+b)log10}//

(log a x)’=1/(xloga)

(loga x)’=(log x/log a)’={(1/log a)(log x)}’=(1/log a)(1/x)=1/(x log a) //

(logxa)’=(loga/logx)’={0loga(logx)’}/(logx)2=loga(1/x)}/(logx)2

           =loga/{x(logx)2}//

(log2x)’ =(log x/log 2)’=1/(x log 2) //         

{log2(12x)}’={log(12x)/log2}’=(12x)’/{(12x)log2}               

             =2/{(12x)log2}=2/{(2x1)log2} // log2を残す)

(log23x)’=(3x)’/(3x log2)=1/(x log2) //      ・{log(ax+b)}’=(ax+b)’/(ax+b)

{log10|3x2|}’=(3x2)’/{(3x2)log10}=3/{(3x2)log10}//(定数のlog10を残す)

(log2x)’=(2x)’/(2x)=2/(2x)=1/x//         (log4x)’=(4x)’/(4x)=4/(4x)=1/x// 

(xlogx)’=x’logx+x(logx)’=logx+1//

(x2logx)’=(x2)’logx + x2(logx)’=2xlogx+x2(1/x)=2xlogx+x//

{x2(logx)3}’=( x2)’(logx)3+ x2{(logx)3}’=2x(logx)3+ x23(logx)2(1/x)

                =2x(logx)3+3x(logx)2//

(logx2)’=(2logx)’=2(1/x)=2/x//または(logx2)’=(x2)’/(x2)=(2 x) /(x2)=2/x//

   ・{(logx)2}’=2logx(logx)’=(2logx)/x//

{(logx)3}’=3(logx)2(logx)’=3(logx)2(1/x)= {3(logx)2}/x//                          

{(logx)x}’ (x>1): y=(logx)x   x>1より(logx)x >0 

    両辺の自然対数をとるとlogy=log{(logx)x}=xlog(logx) 

    両辺をxで微分するとy’/y=1log(logx)+x(1/logx)(1/x)= log{logx}+(1/logx)

                y’=y{log(logx)+(1/logx)}=(logx)x {log(logx)+(1/logx)}//

{log(x2+1)}’=(x2+1)’/(x2+1)=2x/(x2+1) //

{log(x+1)}’={log(x+1)1/2}’={(1/2)log(x+1)}’=(x+1)’/{2(x+1)}=1/{2(x+1)}//

[log{x+(x2+1)}]’={x+(x2+1)}’/ {x+(x2+1)}= 1+{(x2+1)}’/ {x+(x2+1)}

=[1+{2x/(2x2+1)}] /{x+(x2+1)}={(x2+1)+x}/ [{x+(x2+1)}・√(x2+1)]

={x+(x2+1)}/ [{x+(x2+1)}・√(x2+1)]=1/{(x2+1)}//

〔別解①〕[log{x+(x2+1)}]’={x+(x2+1)}’/ {x+(x2+1)}

= 1+{(x2+1)}’/ {x+(x2+1)}

=[1+{2x/(2x2+1)}] /{x+(x2+1)}

=[1+{x/(x2+1)}] /{x+(x2+1)}  分子を√(x2+1)で通分すると

=[{(x2+1)+x}/{(x2+1)}]×[1/{x+(x2+1)}]

=1/{(x2+1)} //   (体系数学Ⅵ)

〔別解②〕{(x2+1)1/2}’=(x2+1)’/{2(x2+1)}=2x/{2(x2+1)}=x/{(x2+1)}, よって

  [log{x+(x2+1)}]’=[1+{x/(x2+1)}]/ {x+(x2+1)}

={(x2+1)+x}/ [{(x2+1)}{x+(x2+1)}]= 1/{(x2+1)} //  推奨

{log(ax+b)’=(ax+b)’/(ax+b)を使う。

{(x2+2x)logx}’=(x2+2x)’logx+(x2+2x)(logx)’=(2x+2)logx+(x2+2x)/x

                =(2x+2)logx+(x+2)//

(log|x1|)’=(x1)’/(x1)=1/(x1) //                                    

(log|x21|)’=(x21)’/(x21)=2x/(x21) //

(log|logx|)’=1/logx(logx)’=(logx)’/ logx =(1/logx)(1/x)=1/(xlogx)//

{log(logx)}’=1/logx(logx)’=(logx)’/ logx =(1/x)/ logx =1/(xlogx)//

{loga|(x1)/(x+1)|}’={loga(|x1|/|x+1|)}’=loga|x1|loga|x+1|

     =[(x1)’/{(x1)loga}][(x+1)’/{(x+1)loga}]

     ={ x+1(x1)}/{( x1)( x+1)loga}=2/{(x21)loga}//

    ☞(logax)’=1/(xloga)に注意

{(x e)log(x2+1)}’=(x e)’log(x2+1)+(x e){log(x2+1)}’

     =ex e1log(x2+1)+ (x e)(x2+1)’/(x2+1)= ex e1log(x2+1)+ (x e)2x/(x2+1)    

     = ex e1log(x2+1)+(2x e+1)/(x2+1)//  ☞{log(ax+b)}’=(ax+b)’/(ax+b)   

{log 3(x2+1)}’=(x2+1)’/{(x2+1)log3}=(2 x) /{(x2+1)log3}//             

{log 3(2x1)}’=(2x1)’/{(2x1)log3}=2/{(2x1)log3}//           

{log(x2+1)}’={(1/2)log(x2+1)}’=(x2+1)’/{2(x2+1)}=(2x)/{2(x2+1)}= x/(x2+1)//

[log{ex(1x)}]’=[logex+log(1x)]’={(ex)’/ex}+{(1x)’/(1x)}=1+{1/(1x)}

     =(1x1)/(1x)=x/(x1)//

[log{( x21)/(x2+1)}]’:   ☞y= log{( x21)/(x2+1)}=log{( x21)/(x2+1)}1/2

           =(1/2){log( x21)log(x2+1)}より                                      

y’=(1/2)[{ ( x21)’/ ( x21)}{(x2+1)’/ (x2+1)}] 通分すると

   =(1/2){2x(x2+1)2x ( x21)}/{( x21)(x2+1)}=2x/(x41)//

[log{x+(x2+4)}]’={x+(x2+4)}’/ {x+(x2+4)}

   ここで{x+(x2+4)}’=1+(1/2)(x2+4)1/2(2x)=1+{2x/2(x2+4)}

     ={(x2+4)+x}/{(x2+4)}                                                       

    よってy’=1/{x+(x2+4)}{x+(x2+4)}/{(x2+4)}=1/(x2+4)//

   ・{3(x+1)logx}’={3(x+1)}’logx+{3(x+1)}(logx)’

=[(x+1)’/{33(x+1)2}logx+{3(x+1)}(1/x)={xlogx+3(x+1)}/{3x3(x+1)2}//

f(x)={log(logx)}, f(a)=0のとき、log{f ’(a)}の値:

f ’(x)=(logx)’/ logx =(1/x)/ logx =1/(xlogx) ①  

f(a)=0よりf(a)=loge(loga)=0 loga=e0=1, a=e1=e a=e ②       

① ②よりlog{f ’(a)}=log{f ’(e)}=log{1/(eloge)}=log{1/e1}=log(1/e)=loge1=1//

{log(sin x)}’=(sin x)’/sin x=cos x/sin x=1/tan x //

{log(cos x)}’=(cos x)’/cos x=(sin x)/cos x=tan x //                           

{log(tanx)}’=(tanx)’/tanx=(1/cos2x)/tanx=(1/cos2x)/(sinx/cosx)

                =cosx/sinx=1/(sinxcosx)//

{log(sin2x)}’=(sin2x)’/(sin2x)={2(sinx)(sinx)’}/(sin2x)=(2cosx)/(sinx)=2/tanx//

[{log(x2+3)}/ (log2x)]’=[{log(x2+3)}’(log2x){log(x2+3)}(log2x)’]/(log2x)’  

=[{(x2+3)’/(x2+3)}(log2x)log(x2+3)(2x)’/(2x)]/(log2x)2                    

=[{(2x)/(x2+3)}(log2x)log(x2+3)2/(2x)]/(log2x)2

=[2 x2(log2x)(x3+3)log(x2+3)]/{x(x2+3)(log2x)}//

   ・([log{(x)+1}]2)’: t=log{(x)+1}とおくと y=t2 よってdy/dt=2t,

dt/dx={(x)+1}’/{(x)+1}=1/[2(x){(x)+1}]=1/[2{x+x}]:

よって(dy/dt)(dt/dx)=(2t)/[1/{2(x+x)}]=[log{(x)+1}]/(x+x)//

   ・[log{(x2a2)/(x2+a2)}]’ (aは定数でa>0, a1): y=log(x2a2)log(x2+a2):

       y’={( x2a2)’/( x2a2)}{( x2+a2)’/( x2+a2)}={2x/( x2a2)}{2x/( x2+a2)}

=2x{(x2+a2)(x2a2)}/{(x2a2)(x2+a2)}=(2ax)/{(x2a2)(x2+a2)}//

   ・{logasinx}’(aは定数でa>0, a1): {logc(ax+b)}’=(ax+b)’/(ax+b)logcより

{logasinx}’=(sinx)’/{(sinx)loga}=cosx/{(sinx)loga}//

   ・[loga{x+(x2a2)}]’ (aは定数でa>0, a1):

            {logc(ax+b)}’=(ax+b)’/(ax+b)logcより

y’={ x+(x2a2)}’/[{x+(x2a2)}loga]: 

ここで分子=1+[(x2a2)’/{2(x2a2)}]=1+[2x/{2(x2a2)}]

        ={2(x2a2)+2x}/{2(x2a2)}={x+(x2a2)}/{(x2a2)}:

よってy’=[{x+(x2a2)}/{(x2a2)}](1/[{x+(x2a2)}loga])

                               =1/{(x2a2)(loga)}//