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

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

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

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

derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int_{0}^{1} (x - \frac{1}{2})^{2} d x

\frac{d y}{d t} - 2 y = 3 e^{- 2 t}

\frac{d}{d x} y = (3 x + 2) (5 x^{2} + 4)

\int \frac{x}{x - 1 + 1} d x

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