derivative of 4/(sin(x+cos(x)))

Lösung

\frac{d}{d x} (\frac{4}{sin ( x ) + cos ( x )})

- \frac{4 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

Schritte zur Lösung

\frac{d}{d x} (\frac{4}{sin ( x ) + cos ( x )})

Entferne die Konstante:

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

= 4 \frac{d}{d x} (\frac{1}{sin ( x ) + cos ( x )})

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

= 4 \frac{d}{d x} ((sin (x) + cos (x))^{- 1})

Wende die Kettenregel an:

- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} \frac{d}{d x} (sin (x) + cos (x))

= - \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} \frac{d}{d x} (sin (x) + cos (x))

\frac{d}{d x} (sin (x) + cos (x)) = cos (x) - sin (x)

= 4 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x)))

Vereinfache

4 (- \frac{1}{( sin ( x ) + cos ( x ) ) ^{2}} (cos (x) - sin (x))) : - \frac{4 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

= - \frac{4 ( cos ( x ) - sin ( x ) )}{1 + sin ( 2 x )}

\int cos (x) - 9 sin (x) d x

\int \frac{5}{x + 5 + x} d x

a re a f (x) = x + 3, [0, 9]

x \to 2 \cdot π lim (ln (cos (x)))

in v erse l a pl a ce \frac{s + 3}{s ^{2} + 5 s + 1}