derivative of 2cos(x+1)

Lösung

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

- 2 sin (x)

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

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

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

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

\frac{d}{d x} (1) = 0

= - 2 sin (x) + 0

Vereinfache

= - 2 sin (x)

\int \frac{sin ( 2 χ + 1 )}{( cos ( 2 χ + 1 ) ) ^{2}} d χ

\int_{2}^{10} \frac{1}{x - 1} d x

\int \frac{1}{16 + 4 x - 2 x ^{2}} d x

\int_{1}^{6} 2 π x (- \frac{3}{2} x + 9) d x

t an g e n t \frac{x + 1}{x + 2}