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

Lösung

ableitung von

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

ln^{2} (2 t) + 2 ln (2 t)

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = (ln (2 t))^{2}

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

\frac{d t}{d t} = 1

\frac{d}{d t} ((ln (2 t))^{2}) = \frac{2 ln ( 2 t )}{t}

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

Vereinfache

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

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

t a y l or \frac{1}{3 - x}

\int \frac{- 1}{2} cos (2 x) d x

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

\frac{\partial}{\partial x} (14 x y e^{- x^{2} - y^{2}})

t an g e n t f (x) = x + 3, at x = 6