What is the derivative of 15e^{-4x}-60xe^{-4x} ?

The derivative of 15e^{-4x}-60xe^{-4x} is -120e^{-4x}+240e^{-4x}x

What is the first derivative of 15e^{-4x}-60xe^{-4x} ?

The first derivative of 15e^{-4x}-60xe^{-4x} is -120e^{-4x}+240e^{-4x}x

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\frac{d}{d x} (15 e^{- 4 x} - 60 x e^{- 4 x})

- 120 e^{- 4 x} + 240 e^{- 4 x} x

Solution steps

\frac{d}{d x} (15 e^{- 4 x} - 60 x e^{- 4 x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (15 e^{- 4 x}) - \frac{d}{d x} (60 x e^{- 4 x})

\frac{d}{d x} (15 e^{- 4 x}) = - 60 e^{- 4 x}

\frac{d}{d x} (60 x e^{- 4 x}) = 60 (e^{- 4 x} - 4 e^{- 4 x} x)

= - 60 e^{- 4 x} - 60 (e^{- 4 x} - 4 e^{- 4 x} x)

Simplify

- 60 e^{- 4 x} - 60 (e^{- 4 x} - 4 e^{- 4 x} x) : - 120 e^{- 4 x} + 240 e^{- 4 x} x

= - 120 e^{- 4 x} + 240 e^{- 4 x} x

y^{^{''}} + y^{^{'}} - 2 y = x^{2}

f (x) = \frac{9}{5 x}

\int (x - 7)^{3} d x

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

y^{^{''}} + 2 y^{^{'}} = 3 + 4 sin (2 x)

What is the derivative of 15e^{-4x}-60xe^{-4x} ?
The derivative of 15e^{-4x}-60xe^{-4x} is -120e^{-4x}+240e^{-4x}x
What is the first derivative of 15e^{-4x}-60xe^{-4x} ?
The first derivative of 15e^{-4x}-60xe^{-4x} is -120e^{-4x}+240e^{-4x}x