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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{1}{x ( 10 - x )} d x

\frac{\partial}{\partial x} (\frac{x + 2}{y + 1})

\frac{d}{d x} (x e^{9 x})

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 9)

\int \frac{( a ^{x} - b ^{x} ) ^{2}}{a ^{x} b ^{x}} d x

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