What is the derivative of t^2-8t+19 ?

The derivative of t^2-8t+19 is 2t-8

What is the first derivative of t^2-8t+19 ?

The first derivative of t^2-8t+19 is 2t-8

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derivative of

t^{2} - 8 t + 19

2 t - 8

Solution steps

\frac{d}{d t} (t^{2} - 8 t + 19)

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} (19)

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

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

\frac{d}{d t} (19) = 0

= 2 t - 8 + 0

Simplify

= 2 t - 8

n = 0 \sum \infty \frac{x ^{n}}{5 ^{n}}

\frac{\partial}{\partial x} (\frac{ln ^{^{'}} ( x ^{2} + 5 )}{y ^{2}})

\frac{\partial}{\partial z} (x y arcsin (yz))

\frac{d}{d x} (3 + \frac{1}{2} cosh (2 x))

f (x) = x^{2} - 9 x + 17, at x = 4

What is the derivative of t^2-8t+19 ?
The derivative of t^2-8t+19 is 2t-8
What is the first derivative of t^2-8t+19 ?
The first derivative of t^2-8t+19 is 2t-8