derivative of 2sin(2x-2cos(2x)-1/5 e^x)

Lösung

\frac{d}{d x} (2 sin (2 x) - 2 cos (2 x) - \frac{1}{5} e^{x})

4 cos (2 x) + 4 sin (2 x) - \frac{1}{5} e^{x}

Schritte zur Lösung

\frac{d}{d x} (2 sin (2 x) - 2 cos (2 x) - \frac{1}{5} e^{x})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (2 sin (2 x)) - \frac{d}{d x} (2 cos (2 x)) - \frac{d}{d x} (\frac{1}{5} e^{x})

\frac{d}{d x} (2 sin (2 x)) = 4 cos (2 x)

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

\frac{d}{d x} (\frac{1}{5} e^{x}) = \frac{1}{5} e^{x}

= 4 cos (2 x) - (- 4 sin (2 x)) - \frac{1}{5} e^{x}

Vereinfache

= 4 cos (2 x) + 4 sin (2 x) - \frac{1}{5} e^{x}

y = - 80 + 208

\frac{\partial}{\partial y} (tan (x) y^{2.5})

d er i v a t i v e f (x) = - \frac{1}{3} t^{3} + 64 t + 3000

\int \frac{( x - 7 )}{x ^{2} - 18 x + 82} d x

\int sin (3 x) tan (3 x) d x