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

The derivative of 1/(3x^2+5x) is -(6x+5)/((3x^2+5x)^2)

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

The first derivative of 1/(3x^2+5x) is -(6x+5)/((3x^2+5x)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

- \frac{6 x + 5}{( 3 x ^{2} + 5 x ) ^{2}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{2} + 5 x) = 6 x + 5

= - \frac{1}{( 3 x ^{2} + 5 x ) ^{2}} (6 x + 5)

Simplify

- \frac{1}{( 3 x ^{2} + 5 x ) ^{2}} (6 x + 5) : - \frac{6 x + 5}{( 3 x ^{2} + 5 x ) ^{2}}

= - \frac{6 x + 5}{( 3 x ^{2} + 5 x ) ^{2}}

\int e^{- y^{2}} y d y

y^{^{'}} = 5 - y

\int_{0}^{5} \frac{x}{1 + 4 x} d x

\int \frac{2 x ^{3} - 3 x ^{2} - 3 x + 2}{x - 2} d x

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

What is the derivative of 1/(3x^2+5x) ?
The derivative of 1/(3x^2+5x) is -(6x+5)/((3x^2+5x)^2)
What is the first derivative of 1/(3x^2+5x) ?
The first derivative of 1/(3x^2+5x) is -(6x+5)/((3x^2+5x)^2)