derivative of (x^4+1sech(2ln(x)))

解

\frac{d}{d x} ((x^{4} + 1) sech (2 ln (x)))

4 x^{3} sech (2 ln (x)) - \frac{2 sech ( 2 ln ( x ) ) tanh ( 2 ln ( x ) ) ( x ^{4} + 1 )}{x}

解答ステップ

\frac{d}{d x} ((x^{4} + 1) sech (2 ln (x)))

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{4} + 1, g = sech (2 ln (x))

= \frac{d}{d x} (x^{4} + 1) sech (2 ln (x)) + \frac{d}{d x} (sech (2 ln (x))) (x^{4} + 1)

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

\frac{d}{d x} (sech (2 ln (x))) = - \frac{2 sech ( 2 ln ( x ) ) tanh ( 2 ln ( x ) )}{x}

= 4 x^{3} sech (2 ln (x)) + (- \frac{2 sech ( 2 ln ( x ) ) tanh ( 2 ln ( x ) )}{x}) (x^{4} + 1)

簡素化

4 x^{3} sech (2 ln (x)) + (- \frac{2 sech ( 2 ln ( x ) ) tanh ( 2 ln ( x ) )}{x}) (x^{4} + 1) : 4 x^{3} sech (2 ln (x)) - \frac{2 sech ( 2 ln ( x ) ) tanh ( 2 ln ( x ) ) ( x ^{4} + 1 )}{x}

= 4 x^{3} sech (2 ln (x)) - \frac{2 sech ( 2 ln ( x ) ) tanh ( 2 ln ( x ) ) ( x ^{4} + 1 )}{x}