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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

y = (5 x + 14)^{5}

(2 + y) d x - 3 (3 - x) d y = 0

\int 18 (200 + t)^{2} d t

x \to - 5 lim (- 9 x^{2} - 6 x - 9)

\frac{d}{d x} (- 5 x)

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