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

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

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

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

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derivative of

\frac{2}{1 + 3 x + x ^{2}}

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(24 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{d y}{d x} = \frac{y ^{3}}{x}, y (1) = 1

\int \frac{x ^{2} - 5 x + 9}{x ^{2} - 5 x + 6} d x

\frac{\partial}{\partial x} (x, y, z)

x^{3}, 0, 3

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