What is the derivative of sqrt(2+xe^{-3x)} ?

The derivative of sqrt(2+xe^{-3x)} is (-3e^{-3x}x+e^{-3x})/(2sqrt(2+e^{-3x)x)}

What is the first derivative of sqrt(2+xe^{-3x)} ?

The first derivative of sqrt(2+xe^{-3x)} is (-3e^{-3x}x+e^{-3x})/(2sqrt(2+e^{-3x)x)}

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

\frac{- 3 e ^{- 3 x} x + e ^{- 3 x}}{2 2 + e ^{- 3 x} x}

Solution steps

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

Apply the chain rule:

\frac{1}{2 2 + x e ^{- 3 x}} \frac{d}{d x} (2 + x e^{- 3 x})

= \frac{1}{2 2 + x e ^{- 3 x}} \frac{d}{d x} (2 + x e^{- 3 x})

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

= \frac{1}{2 2 + x e ^{- 3 x}} (- 3 e^{- 3 x} x + e^{- 3 x})

Simplify

\frac{1}{2 2 + x e ^{- 3 x}} (- 3 e^{- 3 x} x + e^{- 3 x}) : \frac{- 3 e ^{- 3 x} x + e ^{- 3 x}}{2 2 + e ^{- 3 x} x}

= \frac{- 3 e ^{- 3 x} x + e ^{- 3 x}}{2 2 + e ^{- 3 x} x}

\frac{\partial}{\partial x} (x^{\frac{1}{4}} \cdot y^{\frac{1}{2}})

(3 x^{2} + 3 x^{- 2} y) d x + (x^{2} y - x^{- 1}) d y = 0

\int_{0}^{2} π (x^{2} - L)^{2} d x

\frac{2 x + 1}{x + 3}

y = \frac{1}{x}, y = 4 x, y = \frac{1}{4} x

What is the derivative of sqrt(2+xe^{-3x)} ?
The derivative of sqrt(2+xe^{-3x)} is (-3e^{-3x}x+e^{-3x})/(2sqrt(2+e^{-3x)x)}
What is the first derivative of sqrt(2+xe^{-3x)} ?
The first derivative of sqrt(2+xe^{-3x)} is (-3e^{-3x}x+e^{-3x})/(2sqrt(2+e^{-3x)x)}