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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} - y = 11 t e^{2 t}, y (0) = 1

\int 6 x^{2} sin (x) d x

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

\frac{4}{s ^{3} + 4 s ^{2} + 8 s}

\frac{d y}{d t} + 0.5 t y - 4 t = 0

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