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

The derivative of 1/((x^2+1^4)) is -(8x)/((x^2+1)^5)

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

The first derivative of 1/((x^2+1^4)) is -(8x)/((x^2+1)^5)

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

- \frac{8 x}{( x ^{2} + 1 ) ^{5}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{4}{( x ^{2} + 1 ) ^{5}} \frac{d}{d x} (x^{2} + 1)

= - \frac{4}{( x ^{2} + 1 ) ^{5}} \frac{d}{d x} (x^{2} + 1)

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

= - \frac{4}{( x ^{2} + 1 ) ^{5}} \cdot 2 x

Simplify

- \frac{4}{( x ^{2} + 1 ) ^{5}} \cdot 2 x : - \frac{8 x}{( x ^{2} + 1 ) ^{5}}

= - \frac{8 x}{( x ^{2} + 1 ) ^{5}}

\frac{\partial}{\partial y} (- 2 x y)

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

sin (t) = y^{^{'}} + y

n = 1 \sum \infty \frac{- 1}{n + 1}

What is the derivative of 1/((x^2+1^4)) ?
The derivative of 1/((x^2+1^4)) is -(8x)/((x^2+1)^5)
What is the first derivative of 1/((x^2+1^4)) ?
The first derivative of 1/((x^2+1^4)) is -(8x)/((x^2+1)^5)