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

The derivative of 1/(5x^2+6x) is -(10x+6)/((5x^2+6x)^2)

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

The first derivative of 1/(5x^2+6x) is -(10x+6)/((5x^2+6x)^2)

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derivative of

\frac{1}{5 x ^{2} + 6 x}

- \frac{10 x + 6}{( 5 x ^{2} + 6 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{5 x ^{2} + 6 x})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (5 x^{2} + 6 x) = 10 x + 6

= - \frac{1}{( 5 x ^{2} + 6 x ) ^{2}} (10 x + 6)

Simplify

- \frac{1}{( 5 x ^{2} + 6 x ) ^{2}} (10 x + 6) : - \frac{10 x + 6}{( 5 x ^{2} + 6 x ) ^{2}}

= - \frac{10 x + 6}{( 5 x ^{2} + 6 x ) ^{2}}

\frac{\partial}{\partial x} (16 3 y + 2 x)

\frac{\partial}{\partial x} (x e)

\frac{\partial}{\partial y} (y ln (x) + \frac{1}{9} y^{3})

x \to 0 lim (\frac{3 cos ( x )}{x})

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

What is the derivative of 1/(5x^2+6x) ?
The derivative of 1/(5x^2+6x) is -(10x+6)/((5x^2+6x)^2)
What is the first derivative of 1/(5x^2+6x) ?
The first derivative of 1/(5x^2+6x) is -(10x+6)/((5x^2+6x)^2)