What is the derivative of sqrt(x^2+x) ?

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

What is the first derivative of sqrt(x^2+x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 + lim (e^{c s c (x)})

\int e^{3 x} cos (7 x) d x

y^{^{'}} = e^{- t + 3 y}

d er i v a t i v e \frac{x ^{2} + 3 x + 2}{x}

\frac{d}{d x} (tan^{2} (1 + x))

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