derivative of (tan(sqrt(x)-1)/(x^3+2))

Lösung

\frac{d}{d x} (\frac{tan ( x - 1 )}{x ^{3} + 2})

\frac{sec ^{2} ( x - 1 ) ( x ^{3} + 2 ) - 6 x ^{2} x tan ( x - 1 )}{2 x ( x ^{3} + 2 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{tan ( x - 1 )}{x ^{3} + 2})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (tan (x - 1)) = \frac{sec ^{2} ( x - 1 )}{2 x}

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

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

Vereinfache

\frac{\frac{s e c ^{2} ( x - 1 )}{2 x} ( x ^{3} + 2 ) - 3 x ^{2} tan ( x - 1 )}{( x ^{3} + 2 ) ^{2}} : \frac{sec ^{2} ( x - 1 ) ( x ^{3} + 2 ) - 6 x ^{2} x tan ( x - 1 )}{2 x ( x ^{3} + 2 ) ^{2}}

= \frac{sec ^{2} ( x - 1 ) ( x ^{3} + 2 ) - 6 x ^{2} x tan ( x - 1 )}{2 x ( x ^{3} + 2 ) ^{2}}

\int \frac{1}{( 2 x + 3 y ) ^{2}} d x

\int sin^{2} (3 x co) s^{2} 3 x d x

\frac{d y}{d x} = e^{y + x}

\frac{d}{d v} (u sin (v))

\frac{\partial}{\partial y} (ln (5 + x^{2} y^{2}))