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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

3 x^{2} - 2 x

7 (4^{x})

x \to - 5 - lim (\frac{1}{x ^{2} - 25})

\frac{d}{d t} (\frac{5 t}{1 + 2 t ^{2}})

\frac{d y}{d t} = \frac{1}{( y + 1 ) ( t - 2 )}, y (0) = 0

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