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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

t^{2} - 8 t + 16

2 t - 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 t - 8 + 0

Simplify

= 2 t - 8

\frac{d}{d x} ((2 x)^{6 x})

x \to - \infty lim (x^{2} e^{- 3 x})

x - e^{x}

\int (x - 1)^{- 2} d x

\frac{\partial}{\partial x} (arccos (\frac{x}{y}))

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