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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

n \to 0 - lim (5 + \frac{4}{n})

t e^{- 3 t} sin (5 t)

\int \frac{2 x - 1}{x ^{2} + 4} d x

\int \frac{4}{5} x^{3} d x

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