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

The derivative of f(x)= 1/(4x^2+2x) is -(4x+1)/(2x^2(2x+1)^2)

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

The first derivative of f(x)= 1/(4x^2+2x) is -(4x+1)/(2x^2(2x+1)^2)

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derivative of

f (x) = \frac{1}{4 x ^{2} + 2 x}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 x^{2} + 2 x) = 8 x + 2

= - \frac{1}{( 4 x ^{2} + 2 x ) ^{2}} (8 x + 2)

Simplify

- \frac{1}{( 4 x ^{2} + 2 x ) ^{2}} (8 x + 2) : - \frac{4 x + 1}{2 x ^{2} ( 2 x + 1 ) ^{2}}

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

(x^{2} + 2) (x - 5)

\frac{\partial}{\partial x} ((x + 1)^{2} + y^{2} - 1)

x \to 2 lim (\frac{x ^{2} + 2 x - 8}{3 x - 6})

y^{^{'}} + (2 - \frac{x}{x - 1}) y = 0

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