derivative ((5x^4-3x^2-4x+6))/((x-7))

Lösung

ableitung von

\frac{( 5 x ^{4} - 3 x ^{2} - 4 x + 6 )}{( x - 7 )}

\frac{15 x ^{4} - 140 x ^{3} - 3 x ^{2} + 42 x + 22}{( x - 7 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{5 x ^{4} - 3 x ^{2} - 4 x + 6}{x - 7})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 5 x ^{4} - 3 x ^{2} - 4 x + 6 ) ( x - 7 ) - \frac{d}{d x} ( x - 7 ) ( 5 x ^{4} - 3 x ^{2} - 4 x + 6 )}{( x - 7 ) ^{2}}

\frac{d}{d x} (5 x^{4} - 3 x^{2} - 4 x + 6) = 20 x^{3} - 6 x - 4

\frac{d}{d x} (x - 7) = 1

= \frac{( 20 x ^{3} - 6 x - 4 ) ( x - 7 ) - 1 \cdot ( 5 x ^{4} - 3 x ^{2} - 4 x + 6 )}{( x - 7 ) ^{2}}

Vereinfache

\frac{( 20 x ^{3} - 6 x - 4 ) ( x - 7 ) - 1 \cdot ( 5 x ^{4} - 3 x ^{2} - 4 x + 6 )}{( x - 7 ) ^{2}} : \frac{15 x ^{4} - 140 x ^{3} - 3 x ^{2} + 42 x + 22}{( x - 7 ) ^{2}}

= \frac{15 x ^{4} - 140 x ^{3} - 3 x ^{2} + 42 x + 22}{( x - 7 ) ^{2}}

\int (2 x) d x

\int \frac{2}{( 2 x + 4 ) ln ( 2 x + 4 )} d x

\int x^{2} y + tanh (x) d y

x \to 1 lim (1 - 5)

x \to \infty lim (\frac{12}{x})