derivative ((4x^5+4x^4-3x^3))/(x^4)

Lösung

ableitung von

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

\frac{4 x ^{2} + 3}{x ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (4 x^{5} + 4 x^{4} - 3 x^{3}) = 20 x^{4} + 16 x^{3} - 9 x^{2}

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

= \frac{( 20 x ^{4} + 16 x ^{3} - 9 x ^{2} ) x ^{4} - 4 x ^{3} ( 4 x ^{5} + 4 x ^{4} - 3 x ^{3} )}{( x ^{4} ) ^{2}}

Vereinfache

\frac{( 20 x ^{4} + 16 x ^{3} - 9 x ^{2} ) x ^{4} - 4 x ^{3} ( 4 x ^{5} + 4 x ^{4} - 3 x ^{3} )}{( x ^{4} ) ^{2}} : \frac{4 x ^{2} + 3}{x ^{2}}

= \frac{4 x ^{2} + 3}{x ^{2}}

d er i v a t i v e y = 4 + x - 9

y^{3} \frac{d y}{d x} = (y^{4} + 1) cos (x)

t an g e n t f (x) = \frac{6}{x}

\int 3 t d t

l a pl a ce t r an s f or m e^{- 2 t} \cdot (t^{2} + 4 t + 5)