derivative of (i+xj+2xk/(sqrt(5x^2+1)))

解

\frac{d}{d x} (\frac{i + x j + 2 x k}{5 x ^{2} + 1})

\frac{j + 2 k}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}} - i \frac{5 x}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}}

解答ステップ

\frac{d}{d x} (\frac{i + x j + 2 x k}{5 x ^{2} + 1})

簡素化

\frac{i + x j + 2 x k}{5 x ^{2} + 1} : \frac{( j x + 2 k x ) 1 + 5 x ^{2}}{1 + 5 x ^{2}} + i \frac{1 + 5 x ^{2}}{1 + 5 x ^{2}}

= \frac{d}{d x} (\frac{( j x + 2 k x ) 1 + 5 x ^{2}}{1 + 5 x ^{2}} + i \frac{1 + 5 x ^{2}}{1 + 5 x ^{2}})

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{( j x + 2 k x ) 1 + 5 x ^{2}}{1 + 5 x ^{2}}) + \frac{d}{d x} (i \frac{1 + 5 x ^{2}}{1 + 5 x ^{2}})

\frac{d}{d x} (\frac{( j x + 2 k x ) 1 + 5 x ^{2}}{1 + 5 x ^{2}}) = \frac{j + 2 k}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}}

\frac{d}{d x} (i \frac{1 + 5 x ^{2}}{1 + 5 x ^{2}}) = - \frac{5 i x}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}}

= \frac{j + 2 k}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}} - \frac{5 i x}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}}

簡素化

\frac{j + 2 k}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}} - \frac{5 i x}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}} : \frac{j + 2 k}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}} - i \frac{5 x}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}}

= \frac{j + 2 k}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}} - i \frac{5 x}{( 1 + 5 x ^{2} ) ^{\frac{3}{2}}}