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

Lösung

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

- \frac{5}{( 3 x + 2 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Vereinfache

\frac{2 ( 3 x + 2 ) - 3 ( 2 x + 3 )}{( 3 x + 2 ) ^{2}} : - \frac{5}{( 3 x + 2 ) ^{2}}

= - \frac{5}{( 3 x + 2 ) ^{2}}

cos^{2} (x) sin (x) \frac{d y}{d x} + cos^{3} (x) y = 1

t an g e n t x^{3} - 2 x^{2} - 3 x

\frac{\partial}{\partial x} (e^{c o s (x - 2 y)})

in v erse l a pl a ce \frac{( 7 s + 2 )}{s ^{2} + 6 s + 34}

x \to 9 + lim (\frac{7}{x - 9})