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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (1 + 4 x^{2} + x^{4}) = 4 x^{3} + 8 x

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

Simplify

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

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

\int 2 x^{6} - 3 x^{5} + 11 x^{2} + 1 d x

\int (- g t + V) d t

\int e^{9 x} cos (5 x) d x

y = - x^{3} + 2 x

sin (\frac{π}{6} + x)

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