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

Lösung

\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}

Schritte zur Lösung

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

Wende die Produktregel an:

(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)

Vereinfache

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}

x \to 6 + lim (f (x))

y^{^{'}} = (x y^{^{'}} + y)

\frac{d}{d x} (\frac{7 x ( 2 x - 20 ) ^{3}}{4 x + 9})

x \to 0 + lim (x^{\frac{1}{2}} \cdot ln (x))

y^{^{'}} = 4 x + y