derivative f(x)=(5x^3+1)/(2x-3)

Lösung

ableitung von

f (x) = \frac{5 x ^{3} + 1}{2 x - 3}

\frac{20 x ^{3} - 45 x ^{2} - 2}{( 2 x - 3 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{5 x ^{3} + 1}{2 x - 3})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 5 x ^{3} + 1 ) ( 2 x - 3 ) - \frac{d}{d x} ( 2 x - 3 ) ( 5 x ^{3} + 1 )}{( 2 x - 3 ) ^{2}}

\frac{d}{d x} (5 x^{3} + 1) = 15 x^{2}

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

= \frac{15 x ^{2} ( 2 x - 3 ) - 2 ( 5 x ^{3} + 1 )}{( 2 x - 3 ) ^{2}}

Multipliziere aus

15 x^{2} (2 x - 3) - 2 (5 x^{3} + 1) : 20 x^{3} - 45 x^{2} - 2

= \frac{20 x ^{3} - 45 x ^{2} - 2}{( 2 x - 3 ) ^{2}}

\int \frac{5 t - 3}{t + 1} d t

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) !}

\frac{\partial}{\partial x} (2 x e^{x y})

x \to 6 lim (\frac{3 x}{x ^{2} - 6 x})

\int_{1}^{\infty} x d x