What is the derivative of 1/27 (2x-1^3(6-3x)+1) ?

The derivative of 1/27 (2x-1^3(6-3x)+1) is (-(2x-1)^3+6(2x-1)^2(-x+2))/9

What is the first derivative of 1/27 (2x-1^3(6-3x)+1) ?

The first derivative of 1/27 (2x-1^3(6-3x)+1) is (-(2x-1)^3+6(2x-1)^2(-x+2))/9

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\frac{d}{d x} (\frac{1}{27} (2 x - 1)^{3} (6 - 3 x) + 1)

\frac{- ( 2 x - 1 ) ^{3} + 6 ( 2 x - 1 ) ^{2} ( - x + 2 )}{9}

Solution steps

\frac{d}{d x} (\frac{1}{27} (2 x - 1)^{3} (6 - 3 x) + 1)

Simplify

\frac{1}{27} (2 x - 1)^{3} (6 - 3 x) + 1 : \frac{( 2 x - 1 ) ^{3} ( - x + 2 )}{9} + 1

= \frac{d}{d x} (\frac{( 2 x - 1 ) ^{3} ( - x + 2 )}{9} + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{( 2 x - 1 ) ^{3} ( - x + 2 )}{9}) + \frac{d}{d x} (1)

\frac{d}{d x} (\frac{( 2 x - 1 ) ^{3} ( - x + 2 )}{9}) = \frac{6 ( 2 x - 1 ) ^{2} ( - x + 2 ) - ( 2 x - 1 ) ^{3}}{9}

\frac{d}{d x} (1) = 0

= \frac{6 ( 2 x - 1 ) ^{2} ( - x + 2 ) - ( 2 x - 1 ) ^{3}}{9} + 0

Simplify

= \frac{- ( 2 x - 1 ) ^{3} + 6 ( 2 x - 1 ) ^{2} ( - x + 2 )}{9}

\int_{0}^{\frac{π}{4}} sin (4 x) d x

\frac{- 3}{( s - 1 ) ^{2}}

f (x) = \frac{x ^{3}}{3} + 4 x^{2} + 12 x + 19

\int - x cos (x) d x

\frac{\partial}{\partial x} (A e^{- 3 x})

What is the derivative of 1/27 (2x-1^3(6-3x)+1) ?
The derivative of 1/27 (2x-1^3(6-3x)+1) is (-(2x-1)^3+6(2x-1)^2(-x+2))/9
What is the first derivative of 1/27 (2x-1^3(6-3x)+1) ?
The first derivative of 1/27 (2x-1^3(6-3x)+1) is (-(2x-1)^3+6(2x-1)^2(-x+2))/9