derivative of tan^4((1-x/(1+x)))

解

\frac{d}{d x} (tan^{4} (\frac{1 - x}{1 + x}))

- \frac{8 tan ^{3} ( \frac{1 - x}{1 + x} ) sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}

解答ステップ

\frac{d}{d x} (tan^{4} (\frac{1 - x}{1 + x}))

連鎖法則を適用:

4 (tan (\frac{1 - x}{1 + x}))^{3} \frac{d}{d x} (tan (\frac{1 - x}{1 + x}))

= 4 (tan (\frac{1 - x}{1 + x}))^{3} \frac{d}{d x} (tan (\frac{1 - x}{1 + x}))

\frac{d}{d x} (tan (\frac{1 - x}{1 + x})) = - \frac{2 sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}

= 4 (tan (\frac{1 - x}{1 + x}))^{3} (- \frac{2 sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}})

簡素化

4 tan^{3} (\frac{1 - x}{1 + x}) (- \frac{2 sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}) : - \frac{8 tan ^{3} ( \frac{1 - x}{1 + x} ) sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}

= - \frac{8 tan ^{3} ( \frac{1 - x}{1 + x} ) sec ^{2} ( \frac{1 - x}{1 + x} )}{( 1 + x ) ^{2}}