What is the derivative of y= 1/((x^2-1)(x^2+3x+3)) ?

The derivative of y= 1/((x^2-1)(x^2+3x+3)) is -(4x^3+9x^2+4x-3)/((x^2-1)^2(x^2+3x+3)^2)

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

The first derivative of y= 1/((x^2-1)(x^2+3x+3)) is -(4x^3+9x^2+4x-3)/((x^2-1)^2(x^2+3x+3)^2)

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derivative of

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

- \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} ((x^{2} - 1) (x^{2} + 3 x + 3)) = 4 x^{3} + 9 x^{2} + 4 x - 3

= - \frac{1}{( ( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 3)

Simplify

- \frac{1}{( ( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 3) : - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

= - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

\frac{d}{d x} (x^{\frac{9}{4}})

f (θ) = 11 θ

f (s) = \frac{s ^{2} + 1}{s ^{2} + 4}

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

x \to 2 lim ((6 - x) - 2)

What is the derivative of y= 1/((x^2-1)(x^2+3x+3)) ?
The derivative of y= 1/((x^2-1)(x^2+3x+3)) is -(4x^3+9x^2+4x-3)/((x^2-1)^2(x^2+3x+3)^2)
What is the first derivative of y= 1/((x^2-1)(x^2+3x+3)) ?
The first derivative of y= 1/((x^2-1)(x^2+3x+3)) is -(4x^3+9x^2+4x-3)/((x^2-1)^2(x^2+3x+3)^2)