What is the derivative of y=x^3-8x^2+4x-32 ?

The derivative of y=x^3-8x^2+4x-32 is 3x^2-16x+4

What is the first derivative of y=x^3-8x^2+4x-32 ?

The first derivative of y=x^3-8x^2+4x-32 is 3x^2-16x+4

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derivative of

y = x^{3} - 8 x^{2} + 4 x - 32

3 x^{2} - 16 x + 4

Solution steps

\frac{d}{d x} (x^{3} - 8 x^{2} + 4 x - 32)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) - \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (4 x) - \frac{d}{d x} (32)

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

\frac{d}{d x} (8 x^{2}) = 16 x

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (32) = 0

= 3 x^{2} - 16 x + 4 - 0

Simplify

= 3 x^{2} - 16 x + 4

What is the derivative of y=x^3-8x^2+4x-32 ?
The derivative of y=x^3-8x^2+4x-32 is 3x^2-16x+4
What is the first derivative of y=x^3-8x^2+4x-32 ?
The first derivative of y=x^3-8x^2+4x-32 is 3x^2-16x+4