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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{4} - x + 1}{x ^{4} + x + 1}) = \frac{6 x ^{4} - 2}{( x ^{4} + x + 1 ) ^{2}}

= \frac{1}{2 \frac{x ^{4} - x + 1}{x ^{4} + x + 1}} \cdot \frac{6 x ^{4} - 2}{( x ^{4} + x + 1 ) ^{2}}

Simplify

\frac{1}{2 \frac{x ^{4} - x + 1}{x ^{4} + x + 1}} \cdot \frac{6 x ^{4} - 2}{( x ^{4} + x + 1 ) ^{2}} : \frac{3 x ^{4} - 1}{x ^{4} - x + 1 ( x ^{4} + x + 1 ) ^{\frac{3}{2}}}

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

\int cos (x^{3} - 1) d x

x \to 2 lim (- x^{2} + 4 x + 2)

\int (- 2 sin (2 x) + 2 cos (2 x)) d x

h \to 0 lim (\frac{cos ( \frac{π}{6} + h ) - \frac{3}{2}}{h})

\int \frac{ln ( x ) + 1}{x ln ( x )} d x

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