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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 4 \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}}

= 4 \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

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

Simplify

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

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

\frac{lo g _{5} ( x )}{6 + x ^{5}}

\frac{\partial}{\partial x} (y cos (y - 2 x))

\int \frac{- y}{y} d y

y^{^{'}} + 2 x (y + 1) = 0

\int_{- 1}^{1} 2 x sin (1 - x^{2}) d x

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