What is the derivative of (2x/(e^{4x)}) ?

The derivative of (2x/(e^{4x)}) is (2(1-4x))/(e^{4x)}

What is the first derivative of (2x/(e^{4x)}) ?

The first derivative of (2x/(e^{4x)}) is (2(1-4x))/(e^{4x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

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

y^{^{''}} + y = sin (6 t), y (0) = 0, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (1 + cos^{4} (x^{4}))

sin (ln (w))

(4 y + 2 t - 5) d t + (6 y + 4 t - 1) d y = 0

What is the derivative of (2x/(e^{4x)}) ?
The derivative of (2x/(e^{4x)}) is (2(1-4x))/(e^{4x)}
What is the first derivative of (2x/(e^{4x)}) ?
The first derivative of (2x/(e^{4x)}) is (2(1-4x))/(e^{4x)}