f^'(x)=sec(2x)tan(2x)-4/(x^2)

Lösung

f^{^{'}} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

f (x) = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + c_{1}

Schritte zur Lösung

f^{'} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

f (x) = \int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} d x

\int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} d x = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + c_{1}

f (x) = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + c_{1}

n = 0 \sum \infty x^{n} \frac{1}{n}

\frac{d}{d x} (2 x^{3} + \frac{45 x ^{2}}{2} + 21 x + 4)

\int x^{2} 3 + x d x

d er i v a t i v e y = \frac{e ^{x} + 1}{e ^{x} - 1}

\frac{\partial}{\partial x} (- \frac{y}{x ^{2}} + 4 x + \frac{1}{y})