derivative (x^{2.3}+4)/(x^{3.5)+1}

Lösung

ableitung von

\frac{x ^{2.3} + 4}{x ^{3.5} + 1}

\frac{- 1.2 x ^{4.8} - 14 x ^{2.5} + 2.3 x ^{1.3}}{( x ^{3.5} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2.3} + 4}{x ^{3.5} + 1})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{2.3} + 4 ) ( x ^{3.5} + 1 ) - \frac{d}{d x} ( x ^{3.5} + 1 ) ( x ^{2.3} + 4 )}{( x ^{3.5} + 1 ) ^{2}}

\frac{d}{d x} (x^{2.3} + 4) = 2.3 x^{1.3}

\frac{d}{d x} (x^{3.5} + 1) = 3.5 x^{2.5}

= \frac{2.3 x ^{1.3} ( x ^{3.5} + 1 ) - 3.5 x ^{2.5} ( x ^{2.3} + 4 )}{( x ^{3.5} + 1 ) ^{2}}

Multipliziere aus

2.3 x^{1.3} (x^{3.5} + 1) - 3.5 x^{2.5} (x^{2.3} + 4) : - 1.2 x^{4.8} - 14 x^{2.5} + 2.3 x^{1.3}

= \frac{- 1.2 x ^{4.8} - 14 x ^{2.5} + 2.3 x ^{1.3}}{( x ^{3.5} + 1 ) ^{2}}

\int w ln (w) d w

y^{^{'}} + \frac{3}{2 x} y = 0

\int_{1}^{5} (4 x - 1) d x

\int ln (x y) d x

\int_{- \infty}^{10} x e^{\frac{x}{5}} d x