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

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

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

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

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derivative of

f (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}}

\int_{0}^{1} 3 x^{2} (x^{3} + 5)^{- 2} d x

\int sin (5 π) t d t

2 y^{^{''}} - 8 y = 0

\frac{d}{d x} (- e^{x} (x - 7))

\frac{\partial}{\partial y} (- 10 y sin (x y))

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