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

The derivative of 1/(4x^2+1) is -(8x)/((4x^2+1)^2)

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

The first derivative of 1/(4x^2+1) is -(8x)/((4x^2+1)^2)

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (\frac{( 2 x + 3 y )}{4 x + 5 y})

f (x) = x^{2} + 5

cos (π x), 1

\int_{0}^{1} \frac{x ^{4} + 1}{x} d x

\int 1 + sinh^{2} (x) d x

What is the derivative of 1/(4x^2+1) ?
The derivative of 1/(4x^2+1) is -(8x)/((4x^2+1)^2)
What is the first derivative of 1/(4x^2+1) ?
The first derivative of 1/(4x^2+1) is -(8x)/((4x^2+1)^2)