derivative of sqrt(1-169x^2)arccos(13x)

Lösung

\frac{d}{d x} (1 - 169 x^{2} arccos (13 x))

- \frac{169 x arccos ( 13 x )}{1 - 169 x ^{2}} - 13

Schritte zur Lösung

\frac{d}{d x} (1 - 169 x^{2} arccos (13 x))

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 1 - 169 x^{2}, g = arccos (13 x)

= \frac{d}{d x} (1 - 169 x^{2}) arccos (13 x) + \frac{d}{d x} (arccos (13 x)) 1 - 169 x^{2}

\frac{d}{d x} (1 - 169 x^{2}) = - \frac{169 x}{1 - 169 x ^{2}}

\frac{d}{d x} (arccos (13 x)) = - \frac{13}{1 - 169 x ^{2}}

= (- \frac{169 x}{1 - 169 x ^{2}}) arccos (13 x) + (- \frac{13}{1 - 169 x ^{2}}) 1 - 169 x^{2}

Vereinfache

(- \frac{169 x}{1 - 169 x ^{2}}) arccos (13 x) + (- \frac{13}{1 - 169 x ^{2}}) 1 - 169 x^{2} : - \frac{169 x arccos ( 13 x )}{1 - 169 x ^{2}} - 13

= - \frac{169 x arccos ( 13 x )}{1 - 169 x ^{2}} - 13

d er i v a t i v e \frac{9 x ^{2} + 9 x}{8 x}

\int \frac{1}{2 - t} + \frac{1}{2 + t} d t

\int_{- \infty}^{4} e^{- x} d x

y^{^{''}} + y^{^{'}} = sin^{2} (x)

a re a x = - 2, x = 1, y = 11 x, y = x^{2} - 12