derivative of (log_{10}(x)^{log_{10}(x)})

解

\frac{d}{d x} ((lo g_{10} (x))^{l o g_{10} (x)})

\frac{lo g _{10} ( x ) ^{l o g_{10} (x)} ( ln ( lo g _{10} ( x ) ) + 1 )}{ln ( 10 ) x}

解答ステップ

\frac{d}{d x} ((lo g_{10} (x))^{l o g_{10} (x)})

指数の規則を適用する:

a^{b} = e^{b l n (a)}

(lo g_{10} (x))^{l o g_{10} (x)} = e^{l o g_{10} (x) l n (l o g_{10} (x))}

= \frac{d}{d x} (e^{l o g_{10} (x) l n (l o g_{10} (x))})

連鎖法則を適用:

e^{l o g_{10} (x) l n (l o g_{10} (x))} \frac{d}{d x} (lo g_{10} (x) ln (lo g_{10} (x)))

= e^{l o g_{10} (x) l n (l o g_{10} (x))} \frac{d}{d x} (lo g_{10} (x) ln (lo g_{10} (x)))

\frac{d}{d x} (lo g_{10} (x) ln (lo g_{10} (x))) = \frac{ln ( lo g _{10} ( x ) ) + 1}{ln ( 10 ) x}

= e^{l o g_{10} (x) l n (l o g_{10} (x))} \frac{ln ( lo g _{10} ( x ) ) + 1}{ln ( 10 ) x}

簡素化

e^{l o g_{10} (x) l n (l o g_{10} (x))} \frac{ln ( lo g _{10} ( x ) ) + 1}{ln ( 10 ) x} : \frac{lo g _{10} ( x ) ^{l o g_{10} (x)} ( ln ( lo g _{10} ( x ) ) + 1 )}{ln ( 10 ) x}

= \frac{lo g _{10} ( x ) ^{l o g_{10} (x)} ( ln ( lo g _{10} ( x ) ) + 1 )}{ln ( 10 ) x}