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

The derivative of 2/(1+4x^2) is -(16x)/((1+4x^2)^2)

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

The first derivative of 2/(1+4x^2) is -(16x)/((1+4x^2)^2)

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

- \frac{16 x}{( 1 + 4 x ^{2} ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

2 (- \frac{1}{( 1 + 4 x ^{2} ) ^{2}} \cdot 8 x) : - \frac{16 x}{( 1 + 4 x ^{2} ) ^{2}}

= - \frac{16 x}{( 1 + 4 x ^{2} ) ^{2}}

\int \frac{x ^{2}}{36 + x ^{2}} d x

x^{3} - 5

\frac{3 s - 1}{s ^{2} - 1}

\frac{\partial}{\partial x} (\frac{( x + y )}{y + z})

\frac{d y}{d x} = \frac{y x}{x - 1}

What is the derivative of 2/(1+4x^2) ?
The derivative of 2/(1+4x^2) is -(16x)/((1+4x^2)^2)
What is the first derivative of 2/(1+4x^2) ?
The first derivative of 2/(1+4x^2) is -(16x)/((1+4x^2)^2)