derivative of (3x^4+1/(x^3))

Lösung

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

\frac{3 ( x ^{4} - 1 )}{x ^{4}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (3 x^{4} + 1) = 12 x^{3}

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

= \frac{12 x ^{3} x ^{3} - 3 x ^{2} ( 3 x ^{4} + 1 )}{( x ^{3} ) ^{2}}

Vereinfache

\frac{12 x ^{3} x ^{3} - 3 x ^{2} ( 3 x ^{4} + 1 )}{( x ^{3} ) ^{2}} : \frac{3 ( x ^{4} - 1 )}{x ^{4}}

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

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

\int \frac{ln ( x )}{x} d x

x^{2} \frac{d ^{2} y}{d x ^{2}} - 5 x y^{^{'}} + 8 y = 4 x^{2}

\int_{0}^{1} 2 π (x + 1) (x^{2}) d x

t an g e n t y = (3)^{x}, at x = 6