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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int e^{- x^{4}} d x

\frac{\partial}{\partial x} (- 2 x y z^{2} + 12)

10 sin (6 t)

\int \frac{x}{x - x ^{2}} d x

k = 2 \sum \infty (\frac{1}{3})^{k}

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