What is the derivative of f(x)=x^2-8x+16 ?

The derivative of f(x)=x^2-8x+16 is 2x-8

What is the first derivative of f(x)=x^2-8x+16 ?

The first derivative of f(x)=x^2-8x+16 is 2x-8

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derivative of

f (x) = 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 (cos (3 x))^{2} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 t} + 25 e^{t} cos (3 t)

0.125 y^{^{''}} + 4 y^{^{'}} + 28 y = 0, y (0) = 0

\int 2 x (x^{2} + 9)^{6} d x

What is the derivative of f(x)=x^2-8x+16 ?
The derivative of f(x)=x^2-8x+16 is 2x-8
What is the first derivative of f(x)=x^2-8x+16 ?
The first derivative of f(x)=x^2-8x+16 is 2x-8