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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{1}{3 x ^{2} + 4 x}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{2} + 4 x) = 6 x + 4

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

Simplify

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

= - \frac{6 x + 4}{( 3 x ^{2} + 4 x ) ^{2}}

\frac{\partial}{\partial x} (\frac{9 x y ^{2}}{x ^{2} + y ^{4}})

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

\int_{0}^{2} \frac{6 x}{( 1 + x ^{2} ) ^{2}} d x

\frac{\partial}{\partial y} (y - x^{3})

\frac{d}{d x} (\frac{( 1 - x )}{x})

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