derivative 2/3 (4y^2+1/2)^{3/2}

Lösung

ableitung von

\frac{2}{3} (4 y^{2} + \frac{1}{2})^{\frac{3}{2}}

4 \cdot 2^{\frac{1}{2}} y (8 y^{2} + 1)^{\frac{1}{2}}

Schritte zur Lösung

\frac{d}{d y} (\frac{2}{3} (4 y^{2} + \frac{1}{2})^{\frac{3}{2}})

Entferne die Konstante:

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

= \frac{2}{3} \frac{d}{d y} ((4 y^{2} + \frac{1}{2})^{\frac{3}{2}})

Wende die Kettenregel an:

\frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \frac{d}{d y} (4 y^{2} + \frac{1}{2})

= \frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \frac{d}{d y} (4 y^{2} + \frac{1}{2})

\frac{d}{d y} (4 y^{2} + \frac{1}{2}) = 8 y

= \frac{2}{3} \cdot \frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \cdot 8 y

Vereinfache

\frac{2}{3} \cdot \frac{3 ( 4 y ^{2} + \frac{1}{2} ) ^{\frac{1}{2}}}{2} \cdot 8 y : 4 \cdot 2^{\frac{1}{2}} y (8 y^{2} + 1)^{\frac{1}{2}}

= 4 \cdot 2^{\frac{1}{2}} y (8 y^{2} + 1)^{\frac{1}{2}}

\frac{d}{d t} (2 t)

25 y^{^{''}} - y = \frac{x e x}{5}

y^{^{'}} + ln (x) y = ln (x)

\frac{d}{d x} (\frac{x - 3}{6 x - 5})

x \to 2 lim (\frac{1}{2 - x})