What is the derivative of f(x)= 1/(4x^2+3x) ?

The derivative of f(x)= 1/(4x^2+3x) is -(8x+3)/((4x^2+3x)^2)

What is the first derivative of f(x)= 1/(4x^2+3x) ?

The first derivative of f(x)= 1/(4x^2+3x) is -(8x+3)/((4x^2+3x)^2)

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derivative of

f (x) = \frac{1}{4 x ^{2} + 3 x}

- \frac{8 x + 3}{( 4 x ^{2} + 3 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{4 x ^{2} + 3 x})

Apply exponent rule:

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

= \frac{d}{d x} ((4 x^{2} + 3 x)^{- 1})

Apply the chain rule:

- \frac{1}{( 4 x ^{2} + 3 x ) ^{2}} \frac{d}{d x} (4 x^{2} + 3 x)

= - \frac{1}{( 4 x ^{2} + 3 x ) ^{2}} \frac{d}{d x} (4 x^{2} + 3 x)

\frac{d}{d x} (4 x^{2} + 3 x) = 8 x + 3

= - \frac{1}{( 4 x ^{2} + 3 x ) ^{2}} (8 x + 3)

Simplify

- \frac{1}{( 4 x ^{2} + 3 x ) ^{2}} (8 x + 3) : - \frac{8 x + 3}{( 4 x ^{2} + 3 x ) ^{2}}

= - \frac{8 x + 3}{( 4 x ^{2} + 3 x ) ^{2}}

\int (2 x + 1) e^{- x} d x

x \to 3 lim (x (x - 3))

y^{^{'''}} - 2 y^{^{'}} - 4 y = 0

t + 1^{2}

\int_{0}^{11} 2 π y (\frac{y ^{2}}{2}) d y

What is the derivative of f(x)= 1/(4x^2+3x) ?
The derivative of f(x)= 1/(4x^2+3x) is -(8x+3)/((4x^2+3x)^2)
What is the first derivative of f(x)= 1/(4x^2+3x) ?
The first derivative of f(x)= 1/(4x^2+3x) is -(8x+3)/((4x^2+3x)^2)