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

The derivative of g(x)=(x-4)^3+3x(x-4)^2 is 12x^2-72x+96

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

The first derivative of g(x)=(x-4)^3+3x(x-4)^2 is 12x^2-72x+96

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derivative of

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

12 x^{2} - 72 x + 96

Solution steps

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

Simplify

(x - 4)^{3} + 3 x (x - 4)^{2} : 4 x^{3} - 36 x^{2} + 96 x - 64

= \frac{d}{d x} (4 x^{3} - 36 x^{2} + 96 x - 64)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{3}) - \frac{d}{d x} (36 x^{2}) + \frac{d}{d x} (96 x) - \frac{d}{d x} (64)

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

\frac{d}{d x} (36 x^{2}) = 72 x

\frac{d}{d x} (96 x) = 96

\frac{d}{d x} (64) = 0

= 12 x^{2} - 72 x + 96 - 0

Simplify

= 12 x^{2} - 72 x + 96

\frac{\partial}{\partial x} (x y e^{- 9 y})

\frac{a}{s}

x = y^{4}, y = - x^{5}

\int_{2}^{6} 7 x^{3} d x

f (x) = e^{x} - 3 x

What is the derivative of g(x)=(x-4)^3+3x(x-4)^2 ?
The derivative of g(x)=(x-4)^3+3x(x-4)^2 is 12x^2-72x+96
What is the first derivative of g(x)=(x-4)^3+3x(x-4)^2 ?
The first derivative of g(x)=(x-4)^3+3x(x-4)^2 is 12x^2-72x+96