derivative of arcsin(4x^5)

Lösung

\frac{d}{d x} (arcsin (4 x^{5}))

\frac{20 x ^{4}}{1 - 16 x ^{10}}

Schritte zur Lösung

\frac{d}{d x} (arcsin (4 x^{5}))

Wende die Kettenregel an:

\frac{1}{1 - ( 4 x ^{5} ) ^{2}} \frac{d}{d x} (4 x^{5})

= \frac{1}{1 - ( 4 x ^{5} ) ^{2}} \frac{d}{d x} (4 x^{5})

\frac{d}{d x} (4 x^{5}) = 20 x^{4}

= \frac{1}{1 - ( 4 x ^{5} ) ^{2}} \cdot 20 x^{4}

Vereinfache

\frac{1}{1 - ( 4 x ^{5} ) ^{2}} \cdot 20 x^{4} : \frac{20 x ^{4}}{1 - 16 x ^{10}}

= \frac{20 x ^{4}}{1 - 16 x ^{10}}

im pl i c i t \frac{d y}{d x}, 9 sin (x) + cos (y) = sin (x) cos (y)

\int_{0}^{1} x e^{10 x} d x

y^{^{''}} - 4 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

d er i v a t i v e f (x) = x^{2} [0.2]

s l o p e (- 10, 8), (- 15, - 7)