derivative of (sqrt(2x^2)/(cos(x)))

Lösung

\frac{d}{d x} (\frac{2 x ^{2}}{cos ( x )})

2 (x sec (x) tan (x) + \frac{x sec ( x )}{∣ x ∣})

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{2}}{cos ( x )})

Vereinfache

\frac{2 x ^{2}}{cos ( x )} : 2 sec (x) x^{2}

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Produktregel an:

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

f = sec (x), g = x^{2}

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

\frac{d}{d x} (sec (x)) = sec (x) tan (x)

\frac{d}{d x} (x^{2}) = \frac{x}{∣ x ∣}

= 2 (sec (x) tan (x) x^{2} + \frac{x}{∣ x ∣} sec (x))

Vereinfache

2 (sec (x) tan (x) x^{2} + \frac{x}{∣ x ∣} sec (x)) : 2 (x sec (x) tan (x) + \frac{x sec ( x )}{∣ x ∣})

= 2 (x sec (x) tan (x) + \frac{x sec ( x )}{∣ x ∣})

\frac{d}{d x} (ln^{4} (x))

in v erse l a pl a ce \frac{s + 2}{s ( s + 1 ) ^{2}}

d er i v a t i v e g (t) = 2 t^{- \frac{3}{4}}

\frac{\partial}{\partial x} (e^{x} + yz)

3 x y^{^{'}} - y (x + 1) + 7 y^{4} = 0