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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

n \to \infty lim (n^{- 2} e^{3 n})

(- 2, - 4), (2, 2)

2 x^{2} + 4 x - 7

\int \frac{1}{4 u ^{2} + 9} d u

\frac{d}{d x} (1 - 3 cos (x))

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