derivative of log_{6}(2x^3-5x^2)

Lösung

\frac{d}{d x} (lo g_{6} (2 x^{3} - 5 x^{2}))

\frac{2 ( 3 x - 5 )}{ln ( 6 ) x ( 2 x - 5 )}

Schritte zur Lösung

\frac{d}{d x} (lo g_{6} (2 x^{3} - 5 x^{2}))

Wende die log Regel an:

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

= \frac{d}{d x} (\frac{ln ( 2 x ^{3} - 5 x ^{2} )}{ln ( 6 )})

Entferne die Konstante:

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

= \frac{1}{ln ( 6 )} \frac{d}{d x} (ln (2 x^{3} - 5 x^{2}))

Wende die Kettenregel an:

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

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

\frac{d}{d x} (2 x^{3} - 5 x^{2}) = 6 x^{2} - 10 x

= \frac{1}{ln ( 6 )} \cdot \frac{1}{2 x ^{3} - 5 x ^{2}} (6 x^{2} - 10 x)

Vereinfache

\frac{1}{ln ( 6 )} \cdot \frac{1}{2 x ^{3} - 5 x ^{2}} (6 x^{2} - 10 x) : \frac{2 ( 3 x - 5 )}{ln ( 6 ) x ( 2 x - 5 )}

= \frac{2 ( 3 x - 5 )}{ln ( 6 ) x ( 2 x - 5 )}

\int tan^{2} (x se) c^{3} x d x

\frac{d}{d x} ((1 + x)^{15})

y^{^{''}} + 4 y = sin (2 t)

\int_{0}^{2} 7^{- x} d x

\int - 10 x d x