derivative log_{3}(x^3+1)

Lösung

ableitung von

lo g_{3} (x^{3} + 1)

\frac{3 x ^{2}}{ln ( 3 ) ( x ^{3} + 1 )}

Schritte zur Lösung

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

Wende die log Regel an:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Kettenregel an:

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

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

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

= \frac{1}{ln ( 3 )} \cdot \frac{1}{x ^{3} + 1} \cdot 3 x^{2}

Vereinfache

\frac{1}{ln ( 3 )} \cdot \frac{1}{x ^{3} + 1} \cdot 3 x^{2} : \frac{3 x ^{2}}{ln ( 3 ) ( x ^{3} + 1 )}

= \frac{3 x ^{2}}{ln ( 3 ) ( x ^{3} + 1 )}

\frac{d}{d x} (\frac{x}{3})

a re a y = 2 x, [4, 9]

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

\frac{d y}{d t} = (y^{2} + 1) t

\frac{d}{d x} (10 + x)