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Beliebt Rechnen >

(\partial)/(\partial x)(Ae^{-b(A,x,v,t)(x-vt)^2})

Lösung

\frac{\partial}{\partial x} (A e^{- b (A, x, v, t) (x - v t)^{2}})

Lösung

A e^{- b (A, x, v, t) (x - v t)^{2}} (- (\frac{\partial}{\partial x} (b (A, x, v, t)) (x - v t)^{2} + 2 (x - v t) b (A, x, v, t)))

Schritte zur Lösung

\frac{\partial}{\partial x} (A e^{- b (A, x, v, t) (x - v t)^{2}})

Behandle

A, v, t

als Konstanten

Entferne die Konstante:

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

= A \frac{\partial}{\partial x} (e^{- b (A, x, v, t) (x - v t)^{2}})

Wende die Kettenregel an:

e^{- b (A, x, v, t) (x - v t)^{2}} \frac{\partial}{\partial x} (- b (A, x, v, t) (x - v t)^{2})

= e^{- b (A, x, v, t) (x - v t)^{2}} \frac{\partial}{\partial x} (- b (A, x, v, t) (x - v t)^{2})

\frac{\partial}{\partial x} (- b (A, x, v, t) (x - v t)^{2}) = - (\frac{\partial}{\partial x} (b (A, x, v, t)) (x - v t)^{2} + 2 (x - v t) b (A, x, v, t))

= A e^{- b (A, x, v, t) (x - v t)^{2}} (- (\frac{\partial}{\partial x} (b (A, x, v, t)) (x - v t)^{2} + 2 (x - v t) b (A, x, v, t)))

Beliebte Beispiele

derivative of (dy/(dx))+25y=0

\frac{d}{d x} (\frac{d y}{d x}) + 25 y = 0

limit as x approaches 0 of (sin^9(x))/x

x \to 0 lim (\frac{sin ^{9} ( x )}{x})

integral of x^2e^{x^3+4}

\int x^{2} e^{x^{3} + 4} d x

derivative f(x)=((x^2-1))/((x^2+1))

d er i v a t i v e f (x) = \frac{( x ^{2} - 1 )}{( x ^{2} + 1 )}

derivative (x+9)/(x-9)

d er i v a t i v e \frac{x + 9}{x - 9}