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

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

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

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

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derivative of

f (x) = ln (x + x^{2} + 1)

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int (7 + e) d x

\int_{1}^{\infty} \frac{ln ( x )}{x} d x

\frac{d}{d x} (x^{4} - \frac{1}{x})

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