derivative of (6x+5/(sqrt(7-3x^2)))

Lösung

\frac{d}{d x} (\frac{6 x + 5}{7 - 3 x ^{2}})

\frac{42 + 15 x}{( 7 - 3 x ^{2} ) 7 - 3 x ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{6 x + 5}{7 - 3 x ^{2}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 6 x + 5 ) 7 - 3 x ^{2} - \frac{d}{d x} ( 7 - 3 x ^{2} ) ( 6 x + 5 )}{( 7 - 3 x ^{2} ) ^{2}}

\frac{d}{d x} (6 x + 5) = 6

\frac{d}{d x} (7 - 3 x^{2}) = - \frac{3 x}{7 - 3 x ^{2}}

= \frac{6 7 - 3 x ^{2} - ( - \frac{3 x}{7 - 3 x ^{2}} ) ( 6 x + 5 )}{( 7 - 3 x ^{2} ) ^{2}}

Vereinfache

\frac{6 7 - 3 x ^{2} - ( - \frac{3 x}{7 - 3 x ^{2}} ) ( 6 x + 5 )}{( 7 - 3 x ^{2} ) ^{2}} : \frac{42 + 15 x}{( 7 - 3 x ^{2} ) 7 - 3 x ^{2}}

= \frac{42 + 15 x}{( 7 - 3 x ^{2} ) 7 - 3 x ^{2}}

\int_{0}^{2 π} (1 + sin (u)) d u

y^{^{''}} = 5 y

\int_{a}^{b} 1 d x

\frac{d r}{d t} = - 2 t r

in v erse l a pl a ce \frac{6}{s ^{2} + 4}