derivative of (arctanh(sqrt(3x^2))^3)

Lösung

\frac{d}{d x} ((arctanh (3 x^{2}))^{3})

\frac{3 3 x arctanh ^{2} ( 3 x )}{∣ x ∣ ( 1 - 3 x ^{2} )}

Schritte zur Lösung

\frac{d}{d x} ((arctanh (3 x^{2}))^{3})

Vereinfache

(arctanh (3 x^{2}))^{3} : arctanh^{3} (3 x^{2})

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

Wende die Kettenregel an:

3 (arctanh (3 x^{2}))^{2} \frac{d}{d x} (arctanh (3 x^{2}))

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

\frac{d}{d x} (arctanh (3 x^{2})) = \frac{3 x}{∣ x ∣ ( 1 - 3 x ^{2} )}

= 3 (arctanh (3 x^{2}))^{2} \frac{3 x}{∣ x ∣ ( 1 - 3 x ^{2} )}

Vereinfache

3 arctanh^{2} (3 x^{2}) \frac{3 x}{∣ x ∣ ( 1 - 3 x ^{2} )} : \frac{3 3 x arctanh ^{2} ( 3 x )}{∣ x ∣ ( 1 - 3 x ^{2} )}

= \frac{3 3 x arctanh ^{2} ( 3 x )}{∣ x ∣ ( 1 - 3 x ^{2} )}

\int x sin (z) d y

t an g e n t f (x) = (x^{3} - 16 x)^{8}, at x = 4

\int_{0}^{\infty} e^{- b x} d x

n = 0 \sum \infty (\frac{3}{4})^{n}

t a y l or x^{4} + x