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

The derivative of (-3x+1)^3(2x-1)^4 is -9(-3x+1)^2(2x-1)^4+8(2x-1)^3(-3x+1)^3

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

The first derivative of (-3x+1)^3(2x-1)^4 is -9(-3x+1)^2(2x-1)^4+8(2x-1)^3(-3x+1)^3

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derivative of

(- 3 x + 1)^{3} (2 x - 1)^{4}

- 9 (- 3 x + 1)^{2} (2 x - 1)^{4} + 8 (2 x - 1)^{3} (- 3 x + 1)^{3}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (- 3 x + 1)^{3}, g = (2 x - 1)^{4}

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

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

\frac{d}{d x} ((2 x - 1)^{4}) = 8 (2 x - 1)^{3}

= (- 9 (- 3 x + 1)^{2}) (2 x - 1)^{4} + 8 (2 x - 1)^{3} (- 3 x + 1)^{3}

Simplify

= - 9 (- 3 x + 1)^{2} (2 x - 1)^{4} + 8 (2 x - 1)^{3} (- 3 x + 1)^{3}

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

x y y^{^{'}} = x^{2} y + y^{2}

\frac{\partial}{\partial y} (y cos (x) + 2 x e^{y})

\frac{\partial}{\partial x} (x^{2} y^{2} + x y + 2)

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

What is the derivative of (-3x+1)^3(2x-1)^4 ?
The derivative of (-3x+1)^3(2x-1)^4 is -9(-3x+1)^2(2x-1)^4+8(2x-1)^3(-3x+1)^3
What is the first derivative of (-3x+1)^3(2x-1)^4 ?
The first derivative of (-3x+1)^3(2x-1)^4 is -9(-3x+1)^2(2x-1)^4+8(2x-1)^3(-3x+1)^3