What is the derivative of 4x^3+4x^2-8x ?

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

What is the first derivative of 4x^3+4x^2-8x ?

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

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

12 x^{2} + 8 x - 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (8 x) = 8

= 12 x^{2} + 8 x - 8

\frac{d}{d x} (ln (x + x^{2} - 11))

f (x) = x + 3

\frac{d y}{d x} = y (x y^{6} + 4)

n = 1 \sum \infty \frac{1}{( 3 n - 1 ) ^{4}}

y^{^{'}} + 5 x y = 0

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