derivative of sec^2(xtan^2(x))

Lösung

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

2 tan^{3} (x) sec^{2} (x) + 2 sec^{4} (x) tan (x)

Schritte zur Lösung

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

Wende die Produktregel an:

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

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

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

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

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

= 2 sec^{2} (x) tan (x) tan^{2} (x) + 2 tan (x) sec^{2} (x) sec^{2} (x)

Vereinfache

2 sec^{2} (x) tan (x) tan^{2} (x) + 2 tan (x) sec^{2} (x) sec^{2} (x) : 2 tan^{3} (x) sec^{2} (x) + 2 sec^{4} (x) tan (x)

= 2 tan^{3} (x) sec^{2} (x) + 2 sec^{4} (x) tan (x)

\int \frac{2 x + 4}{x ^{2} - x} d x

\frac{\partial}{\partial x} (\frac{1}{3} x^{\frac{3}{2}})

\frac{\partial}{\partial y} (6 x cos (x) cos (y))

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

x \to 0 lim \frac{( ( 6 + x ) ^{- 1} - 6 ^{- 1} )}{x}