What is the derivative of (Ax+Be^{-2x}) ?

The derivative of (Ax+Be^{-2x}) is e^{-2x}A-2e^{-2x}(Ax+B)

What is the first derivative of (Ax+Be^{-2x}) ?

The first derivative of (Ax+Be^{-2x}) is e^{-2x}A-2e^{-2x}(Ax+B)

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

e^{- 2 x} A - 2 e^{- 2 x} (A x + B)

Solution steps

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

Treat

A, B

as constants

Apply the Product Rule:

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

f = A x + B, g = e^{- 2 x}

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

\frac{d}{d x} (A x + B) = A

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

= A e^{- 2 x} + (- 2 e^{- 2 x}) (A x + B)

Simplify

= e^{- 2 x} A - 2 e^{- 2 x} (A x + B)

\int \frac{- 5 x - 3}{x ^{2} + 3 x + 4} d x

\frac{\partial}{\partial x} (x e^{w})

x \to 90 lim (tan (x))

\int \frac{e ^{x}}{x ^{4}} d x

y^{^{'}} - 5 y = 2 x^{2} - 5 x + 8

What is the derivative of (Ax+Be^{-2x}) ?
The derivative of (Ax+Be^{-2x}) is e^{-2x}A-2e^{-2x}(Ax+B)
What is the first derivative of (Ax+Be^{-2x}) ?
The first derivative of (Ax+Be^{-2x}) is e^{-2x}A-2e^{-2x}(Ax+B)