derivative y=(8x^{w(y,x)}-5)/(3x^3+7)

Lösung

ableitung von

y = \frac{8 x ^{w (y, x)} - 5}{3 x ^{3} + 7}

\frac{( 8 e ^{w (y, x) l n (x)} ( \frac{d}{d x} ( w ( y , x ) ) ln ( x ) + \frac{1}{x} w ( y , x ) ) - 0 ) ( 3 x ^{3} + 7 ) - 9 x ^{2} ( 8 x ^{w (y, x)} - 5 )}{( 3 x ^{3} + 7 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{8 x ^{w (y, x)} - 5}{3 x ^{3} + 7})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 8 x ^{w (y, x)} - 5 ) ( 3 x ^{3} + 7 ) - \frac{d}{d x} ( 3 x ^{3} + 7 ) ( 8 x ^{w (y, x)} - 5 )}{( 3 x ^{3} + 7 ) ^{2}}

\frac{d}{d x} (8 x^{w (y, x)} - 5) = 8 e^{w (y, x) l n (x)} (\frac{d}{d x} (w (y, x)) ln (x) + \frac{1}{x} w (y, x)) - 0

\frac{d}{d x} (3 x^{3} + 7) = 9 x^{2}

= \frac{( 8 e ^{w (y, x) l n (x)} ( \frac{d}{d x} ( w ( y , x ) ) ln ( x ) + \frac{1}{x} w ( y , x ) ) - 0 ) ( 3 x ^{3} + 7 ) - 9 x ^{2} ( 8 x ^{w (y, x)} - 5 )}{( 3 x ^{3} + 7 ) ^{2}}

\int_{0}^{1} (x^{2} + 9) e^{- x} d x

\int_{0}^{\infty} 12 e^{- 4 x} d x

l a pl a ce t r an s f or m e^{2 t}

\frac{\partial}{\partial y} (- e^{- x} z cos (y))

d er i v a t i v e (x^{2} \cdot 2^{x})^{7}