derivative of (1+csc(x^9)/(1-cot(x^9)))

Lösung

\frac{d}{d x} (\frac{1 + csc ( x ^{9} )}{1 - cot ( x ^{9} )})

\frac{- 9 x ^{8} cot ( x ^{9} ) csc ( x ^{9} ) ( 1 - cot ( x ^{9} ) ) - 9 x ^{8} csc ^{2} ( x ^{9} ) ( 1 + csc ( x ^{9} ) )}{( 1 - cot ( x ^{9} ) ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{1 + csc ( x ^{9} )}{1 - cot ( x ^{9} )})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 1 + csc ( x ^{9} ) ) ( 1 - cot ( x ^{9} ) ) - \frac{d}{d x} ( 1 - cot ( x ^{9} ) ) ( 1 + csc ( x ^{9} ) )}{( 1 - cot ( x ^{9} ) ) ^{2}}

\frac{d}{d x} (1 + csc (x^{9})) = - 9 x^{8} cot (x^{9}) csc (x^{9})

\frac{d}{d x} (1 - cot (x^{9})) = 9 x^{8} csc^{2} (x^{9})

= \frac{( - 9 x ^{8} cot ( x ^{9} ) csc ( x ^{9} ) ) ( 1 - cot ( x ^{9} ) ) - 9 x ^{8} csc ^{2} ( x ^{9} ) ( 1 + csc ( x ^{9} ) )}{( 1 - cot ( x ^{9} ) ) ^{2}}

Vereinfache

= \frac{- 9 x ^{8} cot ( x ^{9} ) csc ( x ^{9} ) ( 1 - cot ( x ^{9} ) ) - 9 x ^{8} csc ^{2} ( x ^{9} ) ( 1 + csc ( x ^{9} ) )}{( 1 - cot ( x ^{9} ) ) ^{2}}

\frac{\partial}{\partial y} (ln (2 w^{3} - 7 y + 3 x y^{2}))

d er i v a t i v e y = \frac{6 x}{1 - tan ( x )}

\frac{\partial}{\partial x} (x^{\frac{1}{4}} \cdot y^{\frac{1}{2}})

\int_{0}^{2} π (x^{2} - L)^{2} d x

\int (x^{6} + 1) d x