(\partial)/(\partial x)((ln^'(x^2+5))/(y^2))

解

\frac{\partial}{\partial x} (\frac{ln ^{^{'}} ( x ^{2} + 5 )}{y ^{2}})

\frac{2' x ln ^{' - 1} ( x ^{2} + 5 )}{y ^{2} ( x ^{2} + 5 )}

解答ステップ

\frac{\partial}{\partial x} (\frac{ln ^{^{'}} ( x ^{2} + 5 )}{y ^{2}})

y

を定数として扱う

定数を除去:

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

= \frac{1}{y ^{2}} \frac{\partial}{\partial x} (ln^{^{^{'}}} (x^{2} + 5))

指数の規則を適用する:

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

ln^{^{'}} (x^{2} + 5) = e^{^{'} l n (l n (x^{2} + 5))}

= \frac{1}{y ^{2}} \frac{\partial}{\partial x} (e^{^{^{'}} l n (l n (x^{2} + 5))})

連鎖法則を適用:

e^{^{'} l n (l n (x^{2} + 5))} \frac{\partial}{\partial x} (^{'} ln (ln (x^{2} + 5)))

= e^{^{'} l n (l n (x^{2} + 5))} \frac{\partial}{\partial x} (^{'} ln (ln (x^{2} + 5)))

\frac{\partial}{\partial x} (^{'} ln (ln (x^{2} + 5))) = \frac{2 ^{'} x}{ln ( x ^{2} + 5 ) ( x ^{2} + 5 )}

= \frac{1}{y ^{2}} e^{^{^{'}} l n (l n (x^{2} + 5))} \frac{2 ^{^{'}} x}{ln ( x ^{2} + 5 ) ( x ^{2} + 5 )}

簡素化

\frac{1}{y ^{2}} e^{^{'} l n (l n (x^{2} + 5))} \frac{2 ^{'} x}{ln ( x ^{2} + 5 ) ( x ^{2} + 5 )} : \frac{2 ^{'} x ln ^{^{'} - 1} ( x ^{2} + 5 )}{y ^{2} ( x ^{2} + 5 )}

= \frac{2 ^{'} x ln ^{^{'} - 1} ( x ^{2} + 5 )}{y ^{2} ( x ^{2} + 5 )}