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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

= - \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}}

\frac{d}{d y} (4 y^{2})

ln (x + x^{2} - 1)

\frac{d}{d x} (6 x + 6 x^{2})

\frac{d v}{d t} = 0.025 \cdot 7 - \frac{v}{2000}, v (0) = 100

\int_{4}^{10} 0.13 x d x

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