derivative of (xtan(x)/(x^2+1))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

d er i v a t i v e f (x) = \frac{1}{2 x + 3}

y^{^{'}} + \frac{2}{x} y = \frac{1}{x ^{2}}

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

y^{^{'}} = 2 (t + 1) y, y (0) = 1

y^{^{''}} + 2 \cdot y^{^{'}} + y = e^{- t}