derivative of ln(2x^4-x^3+3x^2-3x)

Lösung

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

\frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (2 x^{4} - x^{3} + 3 x^{2} - 3 x) = 8 x^{3} - 3 x^{2} + 6 x - 3

= \frac{1}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x} (8 x^{3} - 3 x^{2} + 6 x - 3)

Vereinfache

\frac{1}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x} (8 x^{3} - 3 x^{2} + 6 x - 3) : \frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

= \frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

im pl i c i t \frac{d y}{d x}, x^{3} y^{5} + 3 x = 8 y^{3} + 1

\int_{0}^{4} 1 + e^{- 2 y} d y

\int \frac{28 s + 28}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

\frac{\partial}{\partial x} (3 x + 4 y^{2} + 3 x^{2} y^{3})

\int x - 2 π d x