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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

n = 0 \sum \infty \frac{7}{3 ^{n}}

\frac{d}{d x} (cosh (k x))

\int \frac{1}{tan ^{2} ( u ) + 1} d u

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

\int \frac{2}{3 x} d x

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