derivative of log_{5}(sqrt(4x^2+5x-8))

Lösung

\frac{d}{d x} (lo g_{5} (4 x^{2} + 5 x - 8))

\frac{8 x + 5}{2 ln ( 5 ) ( 4 x ^{2} + 5 x - 8 )}

Schritte zur Lösung

\frac{d}{d x} (lo g_{5} (4 x^{2} + 5 x - 8))

Wende die log Regel an:

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

= \frac{d}{d x} (\frac{ln ( 4 x ^{2} + 5 x - 8 )}{ln ( 5 )})

Entferne die Konstante:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (4 x^{2} + 5 x - 8))

Wende die Kettenregel an:

\frac{1}{4 x ^{2} + 5 x - 8} \frac{d}{d x} (4 x^{2} + 5 x - 8)

= \frac{1}{4 x ^{2} + 5 x - 8} \frac{d}{d x} (4 x^{2} + 5 x - 8)

\frac{d}{d x} (4 x^{2} + 5 x - 8) = \frac{8 x + 5}{2 4 x ^{2} + 5 x - 8}

= \frac{1}{ln ( 5 )} \cdot \frac{1}{4 x ^{2} + 5 x - 8} \cdot \frac{8 x + 5}{2 4 x ^{2} + 5 x - 8}

Vereinfache

\frac{1}{ln ( 5 )} \cdot \frac{1}{4 x ^{2} + 5 x - 8} \cdot \frac{8 x + 5}{2 4 x ^{2} + 5 x - 8} : \frac{8 x + 5}{2 ln ( 5 ) ( 4 x ^{2} + 5 x - 8 )}

= \frac{8 x + 5}{2 ln ( 5 ) ( 4 x ^{2} + 5 x - 8 )}

\frac{\partial}{\partial x} (x y + \frac{y}{x})

\int \frac{cos ^{5} ( 5 x + 1 )}{sin ^{4} ( 5 x + 1 )} d x

t an g e n t y = \frac{( x - 1 )}{x + 1}

d er i v a t i v e - \frac{5}{7 x + 5}

x \to 0 + lim ((1 + \frac{1}{x})^{x})