derivative of ((tan(3x^3+x^2)/(e^x))^3)

Lösung

\frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{3}

\frac{3 tan ^{2} ( 3 x ^{3} + x ^{2} ) ( x sec ^{2} ( x ^{2} ( 1 + 3 x ) ) ( 9 x + 2 ) - tan ( x ^{2} ( 1 + 3 x ) ) )}{e ^{3 x}}

Schritte zur Lösung

\frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{3}

Wende die Kettenregel an:

3 (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{2} \frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})

= 3 (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{2} \frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})

\frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}}) = \frac{x sec ^{2} ( x ^{2} ( 1 + 3 x ) ) ( 9 x + 2 ) - tan ( x ^{2} ( 1 + 3 x ) )}{e ^{x}}

= 3 (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{2} \frac{x sec ^{2} ( x ^{2} ( 1 + 3 x ) ) ( 9 x + 2 ) - tan ( x ^{2} ( 1 + 3 x ) )}{e ^{x}}

Vereinfache

3 (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{2} \frac{x sec ^{2} ( x ^{2} ( 1 + 3 x ) ) ( 9 x + 2 ) - tan ( x ^{2} ( 1 + 3 x ) )}{e ^{x}} : \frac{3 tan ^{2} ( 3 x ^{3} + x ^{2} ) ( x sec ^{2} ( x ^{2} ( 1 + 3 x ) ) ( 9 x + 2 ) - tan ( x ^{2} ( 1 + 3 x ) ) )}{e ^{3 x}}

= \frac{3 tan ^{2} ( 3 x ^{3} + x ^{2} ) ( x sec ^{2} ( x ^{2} ( 1 + 3 x ) ) ( 9 x + 2 ) - tan ( x ^{2} ( 1 + 3 x ) ) )}{e ^{3 x}}