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

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

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

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

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derivative of

\frac{x + 1}{2 x ^{2} + 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{tan ( x )}{sec ( x )})

\frac{\partial}{\partial x} (arcsec (x + yz))

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

\int (tan^{5} (x)) d x

f (x) = e^{x - 3}

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