What is the ((f(x)+x-6)/(4-2x))' ?

The ((f(x)+x-6)/(4-2x))' is (-xf'(x)+f(x)+2f'(x)-4)/(2(x-2)^2)

What is the first ((f(x)+x-6)/(4-2x))' ?

The first ((f(x)+x-6)/(4-2x))' is (-xf'(x)+f(x)+2f'(x)-4)/(2(x-2)^2)

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(\frac{f ( x ) + x - 6}{4 - 2 x})^{'}

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

Solution steps

(\frac{f ( x ) + x - 6}{4 - 2 x})^{'}

Apply the Quotient Rule:

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

= \frac{( f ( x ) + x - 6 ) ^{^{'}} ( 4 - 2 x ) - ( 4 - 2 x ) ^{^{'}} ( f ( x ) + x - 6 )}{( 4 - 2 x ) ^{2}}

(f (x) + x - 6)^{'} = f^{'} (x) + 1

(4 - 2 x)^{'} = - 2

= \frac{( f ^{^{'}} ( x ) + 1 ) ( 4 - 2 x ) - ( - 2 ) ( f ( x ) + x - 6 )}{( 4 - 2 x ) ^{2}}

Simplify

\frac{( f ^{'} ( x ) + 1 ) ( 4 - 2 x ) - ( - 2 ) ( f ( x ) + x - 6 )}{( 4 - 2 x ) ^{2}} : \frac{- x f ^{'} ( x ) + f ( x ) + 2 f ^{'} ( x ) - 4}{2 ( x - 2 ) ^{2}}

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

\int (9 - x^{2}) d x

\frac{d}{d x} (16 ln (x^{2}) - x^{2})

\int \frac{x ^{3}}{x ^{2} + 64} d x

2^{x}

\frac{d}{d x} ((x + ln (x^{2}))^{s i n (x) + 2 c o s (x)})

What is the ((f(x)+x-6)/(4-2x))' ?
The ((f(x)+x-6)/(4-2x))' is (-xf'(x)+f(x)+2f'(x)-4)/(2(x-2)^2)
What is the first ((f(x)+x-6)/(4-2x))' ?
The first ((f(x)+x-6)/(4-2x))' is (-xf'(x)+f(x)+2f'(x)-4)/(2(x-2)^2)