derivative y=((x^7-4x^3+12x))/(x^3)

Lösung

ableitung von

y = \frac{( x ^{7} - 4 x ^{3} + 12 x )}{x ^{3}}

\frac{4 ( x ^{6} - 6 )}{x ^{3}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{7} - 4 x ^{3} + 12 x}{x ^{3}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{7} - 4 x ^{3} + 12 x ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( x ^{7} - 4 x ^{3} + 12 x )}{( x ^{3} ) ^{2}}

\frac{d}{d x} (x^{7} - 4 x^{3} + 12 x) = 7 x^{6} - 12 x^{2} + 12

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

= \frac{( 7 x ^{6} - 12 x ^{2} + 12 ) x ^{3} - 3 x ^{2} ( x ^{7} - 4 x ^{3} + 12 x )}{( x ^{3} ) ^{2}}

Vereinfache

\frac{( 7 x ^{6} - 12 x ^{2} + 12 ) x ^{3} - 3 x ^{2} ( x ^{7} - 4 x ^{3} + 12 x )}{( x ^{3} ) ^{2}} : \frac{4 ( x ^{6} - 6 )}{x ^{3}}

= \frac{4 ( x ^{6} - 6 )}{x ^{3}}

x \to 1 + lim (x^{\frac{9}{1 - x}})

d er i v a t i v e 2 sin (3 π x)

y^{^{'}} + \frac{2}{x} y = x, y (1) = 0

t an g e n t y = e^{x}

p a r i t y x \cdot tan (x)