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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 8 sin^{2} (x) - 2 sin (x) - 1 d x

t an g e n t f (x) = \frac{3}{x}, at x = 1

d er i v a t i v e f (x) = \frac{3 x ^{2}}{x ^{3} - 8}

d er i v a t i v e 4 x^{10 x^{3}}

f (x) = ln (3^{2 - x^{2}})

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