What is the derivative of 3/(1+x^2) ?

The derivative of 3/(1+x^2) is -(6x)/((1+x^2)^2)

What is the first derivative of 3/(1+x^2) ?

The first derivative of 3/(1+x^2) is -(6x)/((1+x^2)^2)

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\frac{d}{d x} (\frac{3}{1 + x ^{2}})

- \frac{6 x}{( 1 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3}{1 + x ^{2}})

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{1}{1 + x ^{2}})

Apply exponent rule:

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

= 3 \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

= 3 (- \frac{1}{( 1 + x ^{2} ) ^{2}} \cdot 2 x)

Simplify

3 (- \frac{1}{( 1 + x ^{2} ) ^{2}} \cdot 2 x) : - \frac{6 x}{( 1 + x ^{2} ) ^{2}}

= - \frac{6 x}{( 1 + x ^{2} ) ^{2}}

x \to 0 lim ((1 - 1 x)^{\frac{1}{x}})

\frac{x}{3 x ^{2} + 4} - \frac{1}{4}

\int \frac{4 x ^{2} - 8 x}{( x - 1 ) ^{2} ( x ^{2} + 1 ) ^{2}} d x

\int sin (x) (1 - cos (x)) d x

\frac{d}{d x} (\frac{x ^{2} - 6 x}{1 + cos ( x )})

What is the derivative of 3/(1+x^2) ?
The derivative of 3/(1+x^2) is -(6x)/((1+x^2)^2)
What is the first derivative of 3/(1+x^2) ?
The first derivative of 3/(1+x^2) is -(6x)/((1+x^2)^2)