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

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

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

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

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derivative of

y = \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}}

3 x^{- 2} - 4 x

e^{1 + (l n (x))^{2}}

1 - y^{2} d x - 1 - x^{2} d y = 0, y (0) = \frac{1}{2}

y^{^{'}} = e^{t - y}

\int (3 x - 11) e^{x} d x

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