What is the derivative of x^2-8x+16 ?

The derivative of x^2-8x+16 is 2x-8

What is the first derivative of x^2-8x+16 ?

The first derivative of x^2-8x+16 is 2x-8

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derivative of

x^{2} - 8 x + 16

2 x - 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (16) = 0

= 2 x - 8 + 0

Simplify

= 2 x - 8

\int_{2}^{\infty} \frac{1}{x ^{3}} d x

y^{^{'}} - 2 y = 3 e^{t}

15000 + 250 0^{t}

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

\int x ln^{2} (d) x d x

What is the derivative of x^2-8x+16 ?
The derivative of x^2-8x+16 is 2x-8
What is the first derivative of x^2-8x+16 ?
The first derivative of x^2-8x+16 is 2x-8