What is the derivative of 1/(4x^2+6x) ?

The derivative of 1/(4x^2+6x) is -(4x+3)/(2x^2(2x+3)^2)

What is the first derivative of 1/(4x^2+6x) ?

The first derivative of 1/(4x^2+6x) is -(4x+3)/(2x^2(2x+3)^2)

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\frac{d}{d x} (\frac{1}{4 x ^{2} + 6 x})

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

Solution steps

\frac{d}{d x} (\frac{1}{4 x ^{2} + 6 x})

Apply exponent rule:

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

= \frac{d}{d x} ((4 x^{2} + 6 x)^{- 1})

Apply the chain rule:

- \frac{1}{( 4 x ^{2} + 6 x ) ^{2}} \frac{d}{d x} (4 x^{2} + 6 x)

= - \frac{1}{( 4 x ^{2} + 6 x ) ^{2}} \frac{d}{d x} (4 x^{2} + 6 x)

\frac{d}{d x} (4 x^{2} + 6 x) = 8 x + 6

= - \frac{1}{( 4 x ^{2} + 6 x ) ^{2}} (8 x + 6)

Simplify

- \frac{1}{( 4 x ^{2} + 6 x ) ^{2}} (8 x + 6) : - \frac{4 x + 3}{2 x ^{2} ( 2 x + 3 ) ^{2}}

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

\frac{4 + x ^{2}}{4 - x}

f (x) = (12 x - 8)^{\frac{1}{2}}, at x = 2

\frac{d y}{d x} = 7 x y^{2}

\int_{0}^{2} \frac{1}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (3 x^{2} y + 2 y)

What is the derivative of 1/(4x^2+6x) ?
The derivative of 1/(4x^2+6x) is -(4x+3)/(2x^2(2x+3)^2)
What is the first derivative of 1/(4x^2+6x) ?
The first derivative of 1/(4x^2+6x) is -(4x+3)/(2x^2(2x+3)^2)