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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int (8^{x} - sec^{2} (x)) d x

x^{2} y^{^{''}} - 2 x y^{^{'}} - 40 y = x^{4}

\int \frac{1}{t ^{2} 225 - t ^{2}} d t

x \to 1 + lim (- 6)

(7 x + 8)^{2}

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