What is the derivative of (3x-4/(12x^2)) ?

The derivative of (3x-4/(12x^2)) is (-3x+8)/(12x^3)

What is the first derivative of (3x-4/(12x^2)) ?

The first derivative of (3x-4/(12x^2)) is (-3x+8)/(12x^3)

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\frac{d}{d x} (\frac{3 x - 4}{12 x ^{2}})

\frac{- 3 x + 8}{12 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{3 x - 4}{12 x ^{2}})

Take the constant out:

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

= \frac{1}{12} \frac{d}{d x} (\frac{3 x - 4}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{1}{12} \cdot \frac{\frac{d}{d x} ( 3 x - 4 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 3 x - 4 )}{( x ^{2} ) ^{2}}

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

\frac{d}{d x} (x^{2}) = 2 x

= \frac{1}{12} \cdot \frac{3 x ^{2} - 2 x ( 3 x - 4 )}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{12} \cdot \frac{3 x ^{2} - 2 x ( 3 x - 4 )}{( x ^{2} ) ^{2}} : \frac{- 3 x + 8}{12 x ^{3}}

= \frac{- 3 x + 8}{12 x ^{3}}

F (x) = x + 1

y^{^{'}} = \frac{( x ^{2} - y ^{2} )}{x y}, y (2) = 2

cos^{4} (x)

f (x) = x^{3} - 4 x, at x = 2

2 x^{2} - 1, 0, 1

What is the derivative of (3x-4/(12x^2)) ?
The derivative of (3x-4/(12x^2)) is (-3x+8)/(12x^3)
What is the first derivative of (3x-4/(12x^2)) ?
The first derivative of (3x-4/(12x^2)) is (-3x+8)/(12x^3)