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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 4 \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}}

= 4 \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

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

Simplify

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

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

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

x \to 1 + lim (\frac{4}{x ^{(1 - x)}})

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

\int_{0}^{\frac{π}{2}} \frac{1}{x ^{2} + 2} d x

x \to \infty lim (\frac{x ^{5}}{x ^{10} + 9})

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