derivative y=ln(t(t^2+3)^3)

Lösung

ableitung von

y = ln (t (t^{2} + 3)^{3})

\frac{7 t ^{2} + 3}{t ( t ^{2} + 3 )}

Schritte zur Lösung

\frac{d}{d t} (ln (t (t^{2} + 3)^{3}))

Wende die Kettenregel an:

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

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

\frac{d}{d t} (t (t^{2} + 3)^{3}) = (t^{2} + 3)^{3} + 6 t^{2} (t^{2} + 3)^{2}

= \frac{1}{t ( t ^{2} + 3 ) ^{3}} ((t^{2} + 3)^{3} + 6 t^{2} (t^{2} + 3)^{2})

Vereinfache

\frac{1}{t ( t ^{2} + 3 ) ^{3}} ((t^{2} + 3)^{3} + 6 t^{2} (t^{2} + 3)^{2}) : \frac{7 t ^{2} + 3}{t ( t ^{2} + 3 )}

= \frac{7 t ^{2} + 3}{t ( t ^{2} + 3 )}

\int \frac{2 x}{x ^{2} - 9} d x

\frac{d}{d x} (- 4 e^{- 2 x})

f (x) = \frac{( tan ( x ) - 1 )}{sec ( x )}

d er i v a t i v e 3^{9 x}

x \to \infty lim ((3^{x} + 2^{x})^{\frac{1}{x}})