derivative of arccsc(sqrt((x+1/(1-x))))

Lösung

\frac{d}{d x} (arccsc (\frac{x + 1}{1 - x}))

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

Schritte zur Lösung

\frac{d}{d x} (arccsc (\frac{x + 1}{1 - x}))

Wende die Kettenregel an:

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

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

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

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

Vereinfache

- \frac{1}{( \frac{x + 1}{1 - x} ) ^{2} ( \frac{x + 1}{1 - x} ) ^{2} - 1} \cdot \frac{1}{x + 1 ( - x + 1 ) ^{\frac{3}{2}}} : - \frac{1}{2 x ( x + 1 ) ( - x + 1 ) ^{\frac{1}{2}}}

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e - 4 x, y (0) = - 5, y^{^{'}} (0) = 1

\int 11 - 10 x - x^{2} d x

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d er i v a t i v e f (x) = 3 x^{5} + 2 x^{2}

\int (4 x^{3} - 3 x^{2}) (4 x^{4} - 4 x^{3})^{5} d x