derivative (x^4-sec(4x^2-2))^{-4}

Lösung

ableitung von

(x^{4} - sec (4 x^{2} - 2))^{- 4}

- \frac{4 ( 4 x ^{3} - 8 x sec ( 4 x ^{2} - 2 ) tan ( 4 x ^{2} - 2 ) )}{( x ^{4} - sec ( 4 x ^{2} - 2 ) ) ^{5}}

Schritte zur Lösung

\frac{d}{d x} ((x^{4} - sec (4 x^{2} - 2))^{- 4})

Wende die Kettenregel an:

- \frac{4}{( x ^{4} - sec ( 4 x ^{2} - 2 ) ) ^{5}} \frac{d}{d x} (x^{4} - sec (4 x^{2} - 2))

= - \frac{4}{( x ^{4} - sec ( 4 x ^{2} - 2 ) ) ^{5}} \frac{d}{d x} (x^{4} - sec (4 x^{2} - 2))

\frac{d}{d x} (x^{4} - sec (4 x^{2} - 2)) = 4 x^{3} - sec (4 x^{2} - 2) tan (4 x^{2} - 2) \cdot 8 x

= - \frac{4}{( x ^{4} - sec ( 4 x ^{2} - 2 ) ) ^{5}} (4 x^{3} - sec (4 x^{2} - 2) tan (4 x^{2} - 2) \cdot 8 x)

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{4 ( 4 x ^{3} - 8 x sec ( 4 x ^{2} - 2 ) tan ( 4 x ^{2} - 2 ) )}{( x ^{4} - sec ( 4 x ^{2} - 2 ) ) ^{5}}

\frac{\partial}{\partial x} (\frac{y}{x ^{2}})

\int_{0}^{10} t e^{- 0.6 t} d t

x^{^{'}} = r x - 12

6 y^{^{''}} + 9 y^{^{'}} - y = 23

\frac{d}{d x} (y cos (x))