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

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

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

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

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derivative of

g (x) = \frac{1}{5 x ^{2} + 2 x}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{''}} + 5 y^{^{'}} + 6.25 y = 0

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

\int_{1}^{4} x^{\frac{1}{2}} ln (x) d x

\frac{1}{s + 40}

\int \frac{1}{2 x + 7} d x

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