derivative of ((1+3x/(3x))(3-x))

Lösung

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

- \frac{x ^{2} + 1}{x ^{2}}

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Produktregel an:

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

f = \frac{1 + 3 x}{x}, g = 3 - x

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

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

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

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

Vereinfache

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

= - \frac{x ^{2} + 1}{x ^{2}}

\int sin^{3} (4 x) cos (4 x) d x

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = 0

h \to 0 lim (\frac{h}{h ^{2} - 7 h + 1})

\frac{d}{d x} (10 0.6 x + 7)

t an g e n t f (x) = x^{2} + 40, at x = 3