derivative ((2x^3+4x^2-2x+3))/((x-3))

Lösung

ableitung von

\frac{( 2 x ^{3} + 4 x ^{2} - 2 x + 3 )}{( x - 3 )}

\frac{4 x ^{3} - 14 x ^{2} - 24 x + 3}{( x - 3 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{3} + 4 x ^{2} - 2 x + 3}{x - 3})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 2 x ^{3} + 4 x ^{2} - 2 x + 3 ) ( x - 3 ) - \frac{d}{d x} ( x - 3 ) ( 2 x ^{3} + 4 x ^{2} - 2 x + 3 )}{( x - 3 ) ^{2}}

\frac{d}{d x} (2 x^{3} + 4 x^{2} - 2 x + 3) = 6 x^{2} + 8 x - 2

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

= \frac{( 6 x ^{2} + 8 x - 2 ) ( x - 3 ) - 1 \cdot ( 2 x ^{3} + 4 x ^{2} - 2 x + 3 )}{( x - 3 ) ^{2}}

Vereinfache

\frac{( 6 x ^{2} + 8 x - 2 ) ( x - 3 ) - 1 \cdot ( 2 x ^{3} + 4 x ^{2} - 2 x + 3 )}{( x - 3 ) ^{2}} : \frac{4 x ^{3} - 14 x ^{2} - 24 x + 3}{( x - 3 ) ^{2}}

= \frac{4 x ^{3} - 14 x ^{2} - 24 x + 3}{( x - 3 ) ^{2}}

\int x + 3 d x

\frac{\partial}{\partial x} (e^{x} y + 5 x y^{2} - x^{2})

n = 0 \sum \infty \frac{3}{n ^{4}}

\frac{d}{d x} (x cos (x) - sin (x))

\int sin (t) cos (n t) d t