(12x+x^2+0.05x^3)^'

Lösung

(12 x + x^{2} + 0.05 x^{3})^{^{'}}

12 + 2 x + 0.15 x^{2}

Schritte zur Lösung

(12 x + x^{2} + 0.05 x^{3})^{'}

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (12 x)^{^{'}} + (x^{2})^{^{'}} + (0.05 x^{3})^{^{'}}

(12 x)^{'} = 12

(x^{2})^{'} = 2 x

(0.05 x^{3})^{'} = 0.15 x^{2}

= 12 + 2 x + 0.15 x^{2}

\int_{- 1}^{1} e^{y} - (y^{2} - 2) d y

n = 2 \sum \infty \frac{3}{4 ^{n}}

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2}))

\frac{\partial}{\partial x} (13.12 + 0.6215 x - 11.37 y^{0.16} + 0.3965 x y^{0.16})

t an g e n t f (x) = sin (x) cos (x)