What is the derivative of sqrt(ln(x^2+1)) ?

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

ln (x^{2} + 1)

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{- 2}^{0} \frac{x}{2} + 1 d x

y^{^{'}} = \frac{x e ^{y}}{y}

t^{2} y^{^{''}} - 4 t y^{^{'}} + 6 y = 0

x \to 0 lim (\frac{2 x}{2 x ^{2}})

\int_{3}^{5} e^{6 x} d x

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