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

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

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

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

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derivative of

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

\frac{2 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) = 2 x

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

Simplify

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

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

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