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

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

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

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

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{x + x}{e ^{x}})

\frac{2 x + 1 - 2 x - 2 x x}{2 e ^{x} x}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{2 x + 1 - 2 x - 2 x x}{2 e ^{x} x}

\int 2 tan (x) sec (x) d x

(16, 13), (- 8, - 17)

\int_{4}^{\infty} e^{- 9 x} d x

\int 16 - 9 x^{2} d x

\frac{\partial}{\partial x} (3 x + y^{2})

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