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

The derivative of f(x)= 4/((x^2+1)^2) is -(16x)/((x^2+1)^3)

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

The first derivative of f(x)= 4/((x^2+1)^2) is -(16x)/((x^2+1)^3)

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derivative of

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

- \frac{16 x}{( x ^{2} + 1 ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= 4 (- \frac{2}{( x ^{2} + 1 ) ^{3}} \cdot 2 x)

Simplify

4 (- \frac{2}{( x ^{2} + 1 ) ^{3}} \cdot 2 x) : - \frac{16 x}{( x ^{2} + 1 ) ^{3}}

= - \frac{16 x}{( x ^{2} + 1 ) ^{3}}

\frac{d}{d x} (6 x - 9)

y^{^{'''}} - 9 y^{^{''}} + 20 y^{^{'}} = 0

\frac{\partial}{\partial x} (\frac{( x + y )}{x y - 1})

\int_{15}^{30} 9 d x

\frac{d}{d x} (\frac{x}{x ^{2} + 36})

What is the derivative of f(x)= 4/((x^2+1)^2) ?
The derivative of f(x)= 4/((x^2+1)^2) is -(16x)/((x^2+1)^3)
What is the first derivative of f(x)= 4/((x^2+1)^2) ?
The first derivative of f(x)= 4/((x^2+1)^2) is -(16x)/((x^2+1)^3)