What is the derivative of 1/((x^2-3)(x^2+3x+5)) ?

The derivative of 1/((x^2-3)(x^2+3x+5)) is -(4x^3+9x^2+4x-9)/((x^2-3)^2(x^2+3x+5)^2)

What is the first derivative of 1/((x^2-3)(x^2+3x+5)) ?

The first derivative of 1/((x^2-3)(x^2+3x+5)) is -(4x^3+9x^2+4x-9)/((x^2-3)^2(x^2+3x+5)^2)

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derivative of

\frac{1}{( x ^{2} - 3 ) ( x ^{2} + 3 x + 5 )}

- \frac{4 x ^{3} + 9 x ^{2} + 4 x - 9}{( x ^{2} - 3 ) ^{2} ( x ^{2} + 3 x + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{( x ^{2} - 3 ) ( x ^{2} + 3 x + 5 )})

Apply exponent rule:

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

= \frac{d}{d x} (((x^{2} - 3) (x^{2} + 3 x + 5))^{- 1})

Apply the chain rule:

- \frac{1}{( ( x ^{2} - 3 ) ( x ^{2} + 3 x + 5 ) ) ^{2}} \frac{d}{d x} ((x^{2} - 3) (x^{2} + 3 x + 5))

= - \frac{1}{( ( x ^{2} - 3 ) ( x ^{2} + 3 x + 5 ) ) ^{2}} \frac{d}{d x} ((x^{2} - 3) (x^{2} + 3 x + 5))

\frac{d}{d x} ((x^{2} - 3) (x^{2} + 3 x + 5)) = 4 x^{3} + 9 x^{2} + 4 x - 9

= - \frac{1}{( ( x ^{2} - 3 ) ( x ^{2} + 3 x + 5 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 9)

Simplify

- \frac{1}{( ( x ^{2} - 3 ) ( x ^{2} + 3 x + 5 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 9) : - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 9}{( x ^{2} - 3 ) ^{2} ( x ^{2} + 3 x + 5 ) ^{2}}

= - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 9}{( x ^{2} - 3 ) ^{2} ( x ^{2} + 3 x + 5 ) ^{2}}

\frac{\partial}{\partial y} (y (y + sin (x)))

\int_{0}^{0.6} \frac{x ^{2}}{9 - 25 x ^{2}} d x

2 x^{4} - 9 x^{2}

x y^{^{'}} + y = 3

t y^{^{'}} + (t + 1) y = t, y (ln (3)) = 1

What is the derivative of 1/((x^2-3)(x^2+3x+5)) ?
The derivative of 1/((x^2-3)(x^2+3x+5)) is -(4x^3+9x^2+4x-9)/((x^2-3)^2(x^2+3x+5)^2)
What is the first derivative of 1/((x^2-3)(x^2+3x+5)) ?
The first derivative of 1/((x^2-3)(x^2+3x+5)) is -(4x^3+9x^2+4x-9)/((x^2-3)^2(x^2+3x+5)^2)