What is the derivative of (c_{1}+c_{2}xe^{-3x}) ?

The derivative of (c_{1}+c_{2}xe^{-3x}) is c_{2}e^{-3x}-3e^{-3x}(c_{1}+c_{2}x)

What is the first derivative of (c_{1}+c_{2}xe^{-3x}) ?

The first derivative of (c_{1}+c_{2}xe^{-3x}) is c_{2}e^{-3x}-3e^{-3x}(c_{1}+c_{2}x)

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

c_{2} e^{- 3 x} - 3 e^{- 3 x} (c_{1} + c_{2} x)

Solution steps

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

Apply the Product Rule:

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

f = c_{1} + c_{2} x, g = e^{- 3 x}

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

\frac{d}{d x} (c_{1} + c_{2} x) = c_{2}

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

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

Simplify

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

x \to 0 lim (\frac{( 1 + x ^{2} ) - 1}{x})

x \to \frac{π}{3} lim (sin (3 x))

n = 5 \sum \infty e^{7 - 6 n}

x \to \infty lim (x^{2} - 3)

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

What is the derivative of (c_{1}+c_{2}xe^{-3x}) ?
The derivative of (c_{1}+c_{2}xe^{-3x}) is c_{2}e^{-3x}-3e^{-3x}(c_{1}+c_{2}x)
What is the first derivative of (c_{1}+c_{2}xe^{-3x}) ?
The first derivative of (c_{1}+c_{2}xe^{-3x}) is c_{2}e^{-3x}-3e^{-3x}(c_{1}+c_{2}x)