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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{1}^{e^{2}} \frac{ln ^{6} ( x ^{2} )}{x} d x

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

\frac{x}{36 + x ^{2}}

x^{2} y d y - x y^{2} d x - x^{3} y^{2} d x = 0

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

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