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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{2} (y - 1) (2 y + 1) d y

- 6 y^{^{''}} + 24 y = 0, y (- 1) = 1, y^{^{'}} (1) = 1

f (x) = \frac{x - 3}{2}

3 e^{- 4 t} - 5 e^{4 t}

\frac{\partial}{\partial y} (\frac{x}{y ^{2} + x ^{2}})

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