derivative of 1-(1-x/(40)^3)

Lösung

\frac{d}{d x} (1 - (1 - \frac{x}{40})^{3})

\frac{3 x ^{2}}{64000} - \frac{3 x}{800} + \frac{3}{40}

Schritte zur Lösung

\frac{d}{d x} (1 - (1 - \frac{x}{40})^{3})

Vereinfache

1 - (1 - \frac{x}{40})^{3} : \frac{x ^{3}}{64000} - \frac{3 x ^{2}}{1600} + \frac{3 x}{40}

= \frac{d}{d x} (\frac{x ^{3}}{64000} - \frac{3 x ^{2}}{1600} + \frac{3 x}{40})

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (\frac{x ^{3}}{64000}) - \frac{d}{d x} (\frac{3 x ^{2}}{1600}) + \frac{d}{d x} (\frac{3 x}{40})

\frac{d}{d x} (\frac{x ^{3}}{64000}) = \frac{3 x ^{2}}{64000}

\frac{d}{d x} (\frac{3 x ^{2}}{1600}) = \frac{3 x}{800}

\frac{d}{d x} (\frac{3 x}{40}) = \frac{3}{40}

= \frac{3 x ^{2}}{64000} - \frac{3 x}{800} + \frac{3}{40}

\frac{d y}{d x} = \frac{x}{x ^{2} + 1}

2 x y^{^{'}} + y = 6 x, y (4) = 14

\int 2 y + 2 d y

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

t an g e n t f (x) = \frac{x ^{2}}{x - 5}