derivative of (x^4-2x^2+5x+1/(x^4))

Lösung

\frac{d}{d x} (\frac{x ^{4} - 2 x ^{2} + 5 x + 1}{x ^{4}})

\frac{4 x ^{2} - 15 x - 4}{x ^{5}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{4} - 2 x ^{2} + 5 x + 1}{x ^{4}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{4} - 2 x ^{2} + 5 x + 1 ) x ^{4} - \frac{d}{d x} ( x ^{4} ) ( x ^{4} - 2 x ^{2} + 5 x + 1 )}{( x ^{4} ) ^{2}}

\frac{d}{d x} (x^{4} - 2 x^{2} + 5 x + 1) = 4 x^{3} - 4 x + 5

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

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

Vereinfache

\frac{( 4 x ^{3} - 4 x + 5 ) x ^{4} - 4 x ^{3} ( x ^{4} - 2 x ^{2} + 5 x + 1 )}{( x ^{4} ) ^{2}} : \frac{4 x ^{2} - 15 x - 4}{x ^{5}}

= \frac{4 x ^{2} - 15 x - 4}{x ^{5}}

xv \frac{d v}{d x} + v^{2} = 32 x

x \to 1 lim (\frac{x - 1}{∣ x - 1 ∣})

\int_{4}^{9} 2 x d x

n = 1 \sum \infty \frac{n ^{2}}{2 n ^{2}}

d er i v a t i v e y = 4 x^{2} sin (x) tan (x)