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

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

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

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

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

- \frac{1}{2 x ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d x}{d x} = 1

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

Simplify

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

= - \frac{1}{2 x ^{2}}

\frac{d}{d x} (- sin (7 x^{2}) \cdot 14 x)

\frac{d x}{d t} + 2 t x = t, x (0) = \frac{3}{2}

f (x) = 5 x^{3} + 4 x, at x = - 3

f (x) = x \cdot cos (x)

\frac{e ^{x}}{8 + e ^{x}}

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