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

The derivative of (2x/(e^x)) is (2(1-x))/(e^x)

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

The first derivative of (2x/(e^x)) is (2(1-x))/(e^x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\frac{x}{x ^{2} + 5}

\int cos^{12} (x) sin^{15} (x) d x

cos (15 t)

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = 0

What is the derivative of (2x/(e^x)) ?
The derivative of (2x/(e^x)) is (2(1-x))/(e^x)
What is the first derivative of (2x/(e^x)) ?
The first derivative of (2x/(e^x)) is (2(1-x))/(e^x)