derivative of (x^2+13x+36/(x^2+8x-9))

Lösung

\frac{d}{d x} (\frac{x ^{2} + 13 x + 36}{x ^{2} + 8 x - 9})

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

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + 13 x + 36}{x ^{2} + 8 x - 9})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{2} + 13 x + 36 ) ( x ^{2} + 8 x - 9 ) - \frac{d}{d x} ( x ^{2} + 8 x - 9 ) ( x ^{2} + 13 x + 36 )}{( x ^{2} + 8 x - 9 ) ^{2}}

\frac{d}{d x} (x^{2} + 13 x + 36) = 2 x + 13

\frac{d}{d x} (x^{2} + 8 x - 9) = 2 x + 8

= \frac{( 2 x + 13 ) ( x ^{2} + 8 x - 9 ) - ( 2 x + 8 ) ( x ^{2} + 13 x + 36 )}{( x ^{2} + 8 x - 9 ) ^{2}}

Vereinfache

\frac{( 2 x + 13 ) ( x ^{2} + 8 x - 9 ) - ( 2 x + 8 ) ( x ^{2} + 13 x + 36 )}{( x ^{2} + 8 x - 9 ) ^{2}} : - \frac{5}{( x - 1 ) ^{2}}

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

d er i v a t i v e 9 x^{\frac{2}{3}}

t an g e n t f (x) = 6 sec (x)

\frac{d}{d x} (\frac{ln ( x )}{x ^{4}})

\frac{d y}{d x} + y = e^{5 x}

3 y^{^{'}} - 9 x^{2} = 0