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

The derivative of sqrt((2x+1)/(4x^2+1)) is (-4x^2-4x+1)/(sqrt(2x+1)(4x^2+1)^{3/2)}

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

The first derivative of sqrt((2x+1)/(4x^2+1)) is (-4x^2-4x+1)/(sqrt(2x+1)(4x^2+1)^{3/2)}

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derivative of

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

\frac{- 4 x ^{2} - 4 x + 1}{2 x + 1 ( 4 x ^{2} + 1 ) ^{\frac{3}{2}}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{2 x + 1}{4 x ^{2} + 1}) = \frac{- 8 x ^{2} - 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}}

= \frac{1}{2 \frac{2 x + 1}{4 x ^{2} + 1}} \cdot \frac{- 8 x ^{2} - 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{2 \frac{2 x + 1}{4 x ^{2} + 1}} \cdot \frac{- 8 x ^{2} - 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} : \frac{- 4 x ^{2} - 4 x + 1}{2 x + 1 ( 4 x ^{2} + 1 ) ^{\frac{3}{2}}}

= \frac{- 4 x ^{2} - 4 x + 1}{2 x + 1 ( 4 x ^{2} + 1 ) ^{\frac{3}{2}}}

\int x cos (a x) d x

\frac{\partial}{\partial x} (z e^{x yz})

\frac{1}{1 - x}

x \cdot \frac{d y}{d x} = y^{2} - y

x \to 2 lim (4 x^{2} + x - 4)

What is the derivative of sqrt((2x+1)/(4x^2+1)) ?
The derivative of sqrt((2x+1)/(4x^2+1)) is (-4x^2-4x+1)/(sqrt(2x+1)(4x^2+1)^{3/2)}
What is the first derivative of sqrt((2x+1)/(4x^2+1)) ?
The first derivative of sqrt((2x+1)/(4x^2+1)) is (-4x^2-4x+1)/(sqrt(2x+1)(4x^2+1)^{3/2)}