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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

n = 1 \sum \infty n (sin (nπ))

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

n = 1 \sum \infty \frac{3 n}{- 2 n + 5}

2 e^{3 t}

3600 (1 - \frac{t}{60})^{2}

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