What is the derivative of y= 1/4 (6t^2+8) ?

The derivative of y= 1/4 (6t^2+8) is 3t

What is the first derivative of y= 1/4 (6t^2+8) ?

The first derivative of y= 1/4 (6t^2+8) is 3t

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derivative of

y = \frac{1}{4} (6 t^{2} + 8)

3 t

Solution steps

\frac{d}{d t} (\frac{1}{4} (6 t^{2} + 8))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d t} (6 t^{2} + 8)

Apply the Sum/Difference Rule:

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

= \frac{1}{4} (\frac{d}{d t} (6 t^{2}) + \frac{d}{d t} (8))

\frac{d}{d t} (6 t^{2}) = 12 t

\frac{d}{d t} (8) = 0

= \frac{1}{4} (12 t + 0)

Simplify

\frac{1}{4} (12 t + 0) : 3 t

= 3 t

\int \frac{1}{6} (e^{6 x}) d x

y = cos (e^{x})

n = 0 \sum \infty \frac{n ^{5}}{2 ^{n}}

f (x) = \frac{8}{x ^{2} + 4}, (2, 1)

x \to \infty lim (2)

What is the derivative of y= 1/4 (6t^2+8) ?
The derivative of y= 1/4 (6t^2+8) is 3t
What is the first derivative of y= 1/4 (6t^2+8) ?
The first derivative of y= 1/4 (6t^2+8) is 3t