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

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

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

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

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derivative of

g (x) = \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} (\frac{1}{y} cos (\frac{x}{y}))

θ \to 0 lim (\frac{1 - cos ( θ )}{sin ( θ )})

\int sec (x) tan (x) - 7 e^{x} d x

x \to 6 lim (3)

y^{^{'}} = 2 x y

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