What is the derivative of 1/(5x^2+4x) ?

The derivative of 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)

What is the first derivative of 1/(5x^2+4x) ?

The first derivative of 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)

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\frac{d}{d x} (\frac{1}{5 x ^{2} + 4 x})

- \frac{10 x + 4}{( 5 x ^{2} + 4 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{5 x ^{2} + 4 x})

Apply exponent rule:

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

= \frac{d}{d x} ((5 x^{2} + 4 x)^{- 1})

Apply the chain rule:

- \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} \frac{d}{d x} (5 x^{2} + 4 x)

= - \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} \frac{d}{d x} (5 x^{2} + 4 x)

\frac{d}{d x} (5 x^{2} + 4 x) = 10 x + 4

= - \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} (10 x + 4)

Simplify

- \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} (10 x + 4) : - \frac{10 x + 4}{( 5 x ^{2} + 4 x ) ^{2}}

= - \frac{10 x + 4}{( 5 x ^{2} + 4 x ) ^{2}}

\int (\frac{y}{3 y + 5}) d y

(40 ln (3 x + 1)) - (\frac{x}{3})

x \to 1 lim ((2 x^{2}) - 5 x + 1)

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

\int e^{x} (1 + x) d x

What is the derivative of 1/(5x^2+4x) ?
The derivative of 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)
What is the first derivative of 1/(5x^2+4x) ?
The first derivative of 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)