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

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

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

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

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derivative of

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

- \frac{3}{2 ( x - 2 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (x e^{4 y} sin (3 z))

\int \frac{2}{x ^{2} + 10 x + 29} d x

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

f (x) = x^{2} - x + 5, at x = 5

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

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