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

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

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

The first derivative of f(x)= 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

f (x) = \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 z} (\frac{( 3 x + y )}{z})

\frac{d y}{d t} = - y + 5

y^{^{''}} + 9 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

\int \frac{y ^{2} - 9}{y} d y

9 x - x^{2}, 2 x, 0, 7

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