What is the derivative of-xe^{-2xy} ?

The derivative of-xe^{-2xy} is -e^{-2xy}+2e^{-2xy}xy+2e^{-2xy}x^2(dy)/(dx)

What is the first derivative of-xe^{-2xy} ?

The first derivative of-xe^{-2xy} is -e^{-2xy}+2e^{-2xy}xy+2e^{-2xy}x^2(dy)/(dx)

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\frac{d}{d x} (- x e^{- 2 x y})

- e^{- 2 x y} + 2 e^{- 2 x y} x y + 2 e^{- 2 x y} x^{2} \frac{d y}{d x}

Solution steps

\frac{d}{d x} (- x e^{- 2 x y})

Take the constant out:

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

= - \frac{d}{d x} (x e^{- 2 x y})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{- 2 x y}

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

\frac{d x}{d x} = 1

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

= - (1 \cdot e^{- 2 x y} + (- 2 e^{- 2 x y} (y + x \frac{d y}{d x})) x)

Simplify

- (1 \cdot e^{- 2 x y} + (- 2 e^{- 2 x y} (y + x \frac{d y}{d x})) x) : - e^{- 2 x y} + 2 e^{- 2 x y} x y + 2 e^{- 2 x y} x^{2} \frac{d y}{d x}

= - e^{- 2 x y} + 2 e^{- 2 x y} x y + 2 e^{- 2 x y} x^{2} \frac{d y}{d x}

0.01 (0.2 t^{2} + 4 t + 64)^{\frac{2}{3}}

x \to 2 lim (3 x + 1)

\int \frac{x - 1}{2 x - x + 1} d x

\frac{d}{d x} (2 x e^{- 2 x})

n = 5 \sum \infty \frac{5 ^{n}}{6 ^{n}}

What is the derivative of-xe^{-2xy} ?
The derivative of-xe^{-2xy} is -e^{-2xy}+2e^{-2xy}xy+2e^{-2xy}x^2(dy)/(dx)
What is the first derivative of-xe^{-2xy} ?
The first derivative of-xe^{-2xy} is -e^{-2xy}+2e^{-2xy}xy+2e^{-2xy}x^2(dy)/(dx)