derivative f(x)=(a^{1/x}+b^{1/x})^x

解

導関数

f (x) = (a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x}

ln (a^{\frac{1}{x}} + b^{\frac{1}{x}}) + \frac{- a ^{\frac{1}{x}} ln ( a ) - b ^{\frac{1}{x}} ln ( b )}{x ( a ^{\frac{1}{x}} + b ^{\frac{1}{x}} )} (a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x}

解答ステップ

\frac{d}{d x} ((a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x})

指数の規則を適用する:

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

(a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x} = e^{x l n (a^{\frac{1}{x}} + b^{\frac{1}{x}})}

= \frac{d}{d x} (e^{x l n (a^{\frac{1}{x}} + b^{\frac{1}{x}})})

連鎖法則を適用:

e^{x l n (a^{\frac{1}{x}} + b^{\frac{1}{x}})} \frac{d}{d x} (x ln (a^{\frac{1}{x}} + b^{\frac{1}{x}}))

= e^{x l n (a^{\frac{1}{x}} + b^{\frac{1}{x}})} \frac{d}{d x} (x ln (a^{\frac{1}{x}} + b^{\frac{1}{x}}))

\frac{d}{d x} (x ln (a^{\frac{1}{x}} + b^{\frac{1}{x}})) = ln (a^{\frac{1}{x}} + b^{\frac{1}{x}}) + \frac{- a ^{\frac{1}{x}} ln ( a ) - b ^{\frac{1}{x}} ln ( b )}{x ( a ^{\frac{1}{x}} + b ^{\frac{1}{x}} )}

= e^{x l n (a^{\frac{1}{x}} + b^{\frac{1}{x}})} ln (a^{\frac{1}{x}} + b^{\frac{1}{x}}) + \frac{- a ^{\frac{1}{x}} ln ( a ) - b ^{\frac{1}{x}} ln ( b )}{x ( a ^{\frac{1}{x}} + b ^{\frac{1}{x}} )}

e^{x l n (a^{\frac{1}{x}} + b^{\frac{1}{x}})} = (a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x}

= ln (a^{\frac{1}{x}} + b^{\frac{1}{x}}) + \frac{- a ^{\frac{1}{x}} ln ( a ) - b ^{\frac{1}{x}} ln ( b )}{x ( a ^{\frac{1}{x}} + b ^{\frac{1}{x}} )} (a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x}