What is the derivative of (2x-4/(3e^x)) ?

The derivative of (2x-4/(3e^x)) is (2(-x+3))/(3e^x)

What is the first derivative of (2x-4/(3e^x)) ?

The first derivative of (2x-4/(3e^x)) is (2(-x+3))/(3e^x)

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\frac{d}{d x} (\frac{2 x - 4}{3 e ^{x}})

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

Solution steps

\frac{d}{d x} (\frac{2 x - 4}{3 e ^{x}})

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (\frac{2 x - 4}{e ^{x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{3} \cdot \frac{\frac{d}{d x} ( 2 x - 4 ) e ^{x} - \frac{d}{d x} ( e ^{x} ) ( 2 x - 4 )}{( e ^{x} ) ^{2}}

\frac{d}{d x} (2 x - 4) = 2

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

= \frac{1}{3} \cdot \frac{2 e ^{x} - e ^{x} ( 2 x - 4 )}{( e ^{x} ) ^{2}}

Simplify

\frac{1}{3} \cdot \frac{2 e ^{x} - e ^{x} ( 2 x - 4 )}{( e ^{x} ) ^{2}} : \frac{2 ( - x + 3 )}{3 e ^{x}}

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

x \to 3 lim (\frac{1}{( x ^{2} - 9 )})

\int 2 x e^{- 2 x^{2}} d x

f (x) = x^{5} \cdot e^{x}

n = 1 \sum \infty \frac{( x + 3 ) ^{n}}{n}

\int 20 + 0.6 x d x

What is the derivative of (2x-4/(3e^x)) ?
The derivative of (2x-4/(3e^x)) is (2(-x+3))/(3e^x)
What is the first derivative of (2x-4/(3e^x)) ?
The first derivative of (2x-4/(3e^x)) is (2(-x+3))/(3e^x)