derivative of 1/(x^2(3x+1))

Lösung

\frac{d}{d x} (\frac{1}{x ^{2} ( 3 x + 1 )})

- \frac{9 x + 2}{x ^{3} ( 3 x + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{x ^{2} ( 3 x + 1 )})

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

- \frac{1}{( x ^{2} ( 3 x + 1 ) ) ^{2}} \frac{d}{d x} (x^{2} (3 x + 1))

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

\frac{d}{d x} (x^{2} (3 x + 1)) = 9 x^{2} + 2 x

= - \frac{1}{( x ^{2} ( 3 x + 1 ) ) ^{2}} (9 x^{2} + 2 x)

Vereinfache

- \frac{1}{( x ^{2} ( 3 x + 1 ) ) ^{2}} (9 x^{2} + 2 x) : - \frac{9 x + 2}{x ^{3} ( 3 x + 1 ) ^{2}}

= - \frac{9 x + 2}{x ^{3} ( 3 x + 1 ) ^{2}}

\frac{\partial}{\partial x} (x y^{2} cos (2 + x^{2} y))

a re a 2 sin (x), [0, π]

t an g e n t f (x) = \frac{5}{2 5 x + 6}, at x = 6

\frac{\partial}{\partial x} (3 x)

y^{^{'}} - 4 y = 7 e^{4 x}