d/(da)(blog_{10}(y/x+(0.6x)/(sqrt(a))))

解

\frac{d}{d a} (b lo g_{10} (\frac{y}{x} + \frac{0.6 x}{a}))

\frac{0.43429 \dots b ( - 0.3 x ^{2} + a ^{\frac{3}{2}} \frac{d}{d a} ( y ) )}{a ( a y + 0.6 x ^{2} )}

解答ステップ

\frac{d}{d a} (b lo g_{10} (\frac{y}{x} + \frac{0.6 x}{a}))

b, x

を定数として扱う

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= b \frac{d}{d a} (lo g_{10} (\frac{y}{x} + \frac{0.6 x}{a}))

対数の規則を適用する:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= b \frac{d}{d a} \frac{ln ( \frac{y}{x} + \frac{0.6 x}{a} )}{ln ( 10 )}

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= b \frac{1}{ln ( 10 )} \frac{d}{d a} (ln (\frac{y}{x} + \frac{0.6 x}{a}))

連鎖法則を適用:

\frac{1}{\frac{y}{x} + \frac{0.6 x}{a}} \frac{d}{d a} (\frac{y}{x} + \frac{0.6 x}{a})

= \frac{1}{\frac{y}{x} + \frac{0.6 x}{a}} \frac{d}{d a} (\frac{y}{x} + \frac{0.6 x}{a})

\frac{d}{d a} (\frac{y}{x} + \frac{0.6 x}{a}) = \frac{\frac{d}{d a} ( y )}{x} - \frac{0.3 x}{a ^{\frac{3}{2}}}

= b \frac{1}{ln ( 10 )} \cdot \frac{1}{\frac{y}{x} + \frac{0.6 x}{a}} (\frac{\frac{d}{d a} ( y )}{x} - \frac{0.3 x}{a ^{\frac{3}{2}}})

簡素化

b \frac{1}{ln ( 10 )} \cdot \frac{1}{\frac{y}{x} + \frac{0.6 x}{a}} (\frac{\frac{d}{d a} ( y )}{x} - \frac{0.3 x}{a ^{\frac{3}{2}}}) : \frac{0.43429 \dots b ( - 0.3 x ^{2} + a ^{\frac{3}{2}} \frac{d}{d a} ( y ) )}{a ( a y + 0.6 x ^{2} )}

= \frac{0.43429 \dots b ( - 0.3 x ^{2} + a ^{\frac{3}{2}} \frac{d}{d a} ( y ) )}{a ( a y + 0.6 x ^{2} )}