What is the derivative of 4/(1+x^2) ?

The derivative of 4/(1+x^2) is -(8x)/((1+x^2)^2)

What is the first derivative of 4/(1+x^2) ?

The first derivative of 4/(1+x^2) is -(8x)/((1+x^2)^2)

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derivative of

\frac{4}{1 + x ^{2}}

- \frac{8 x}{( 1 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4}{1 + x ^{2}})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{1 + x ^{2}})

Apply exponent rule:

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

= 4 \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

= 4 (- \frac{1}{( 1 + x ^{2} ) ^{2}} \cdot 2 x)

Simplify

4 (- \frac{1}{( 1 + x ^{2} ) ^{2}} \cdot 2 x) : - \frac{8 x}{( 1 + x ^{2} ) ^{2}}

= - \frac{8 x}{( 1 + x ^{2} ) ^{2}}

\frac{d}{d x} (\frac{1}{e})

\int \frac{1}{( e ^{2 x} + 4 )} d x

\frac{\partial}{\partial y} (2 x y + y^{4})

\int \frac{- x ^{2}}{2} d x

\int_{- π}^{π} x^{2} d x

What is the derivative of 4/(1+x^2) ?
The derivative of 4/(1+x^2) is -(8x)/((1+x^2)^2)
What is the first derivative of 4/(1+x^2) ?
The first derivative of 4/(1+x^2) is -(8x)/((1+x^2)^2)