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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{4 x - 16}{x ^{2} - 8 x + 5} d x

\int \frac{3 y}{y ^{2} + z ^{2}} d y

(x - 1)^{12}

\int x - x^{2} d x

81 y^{^{''}} - 18 y^{^{'}} + y = 0

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