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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (\frac{4 x ^{3}}{x ^{2} + y ^{2}})

\frac{d y}{d x} = 9 y e^{x + 9}

\frac{\partial}{\partial x} (\frac{1}{2 x})

x \to 8 lim (- (2 x - 8))

\int (7 x^{2} - 8 x) e^{2 x} d x

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