derivative y=5ln(4x^2+1)

Lösung

ableitung von

y = 5 ln (4 x^{2} + 1)

\frac{40 x}{4 x ^{2} + 1}

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Kettenregel an:

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

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

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

= 5 \cdot \frac{1}{4 x ^{2} + 1} \cdot 8 x

Vereinfache

5 \cdot \frac{1}{4 x ^{2} + 1} \cdot 8 x : \frac{40 x}{4 x ^{2} + 1}

= \frac{40 x}{4 x ^{2} + 1}

\int t e^{4 t} d t

\int \frac{8}{( 16 x ^{2} + 1 ) ^{2}} d x

\frac{\partial}{\partial y} (e^{x y z^{6}} y z^{6})

ma c l a u r in 1 - x

y^{^{''}} + 8 y^{^{'}} + 16 y = 1521 e^{2 t} cos (9 t)