derivative f(x)=(2sin^3(x)-5x)^{5/3}

Lösung

ableitung von

f (x) = (2 sin^{3} (x) - 5 x)^{\frac{5}{3}}

\frac{5 ( 6 sin ^{2} ( x ) cos ( x ) - 5 ) ( 2 sin ^{3} ( x ) - 5 x ) ^{\frac{2}{3}}}{3}

Schritte zur Lösung

\frac{d}{d x} ((2 sin^{3} (x) - 5 x)^{\frac{5}{3}})

Wende die Kettenregel an:

\frac{5 ( 2 sin ^{3} ( x ) - 5 x ) ^{\frac{2}{3}}}{3} \frac{d}{d x} (2 sin^{3} (x) - 5 x)

= \frac{5 ( 2 sin ^{3} ( x ) - 5 x ) ^{\frac{2}{3}}}{3} \frac{d}{d x} (2 sin^{3} (x) - 5 x)

\frac{d}{d x} (2 sin^{3} (x) - 5 x) = 6 sin^{2} (x) cos (x) - 5

= \frac{5 ( 2 sin ^{3} ( x ) - 5 x ) ^{\frac{2}{3}}}{3} (6 sin^{2} (x) cos (x) - 5)

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 ( 6 sin ^{2} ( x ) cos ( x ) - 5 ) ( 2 sin ^{3} ( x ) - 5 x ) ^{\frac{2}{3}}}{3}

mi d p o in t (3, - 5), (7, 9)

d er i v a t i v e 2 x \cdot e^{x^{2}}

e x t re m e f (x, y) = - e^{x - y^{2} + x y}

mi d p o in t (15, 3), (2, - 14)

d er i v a t i v e f (x) = x^{2} - 5