What is the derivative of 1/4 (x^4+8) ?

The derivative of 1/4 (x^4+8) is x^3

What is the first derivative of 1/4 (x^4+8) ?

The first derivative of 1/4 (x^4+8) is x^3

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\frac{d}{d x} (\frac{1}{4} (x^{4} + 8))

x^{3}

Solution steps

\frac{d}{d x} (\frac{1}{4} (x^{4} + 8))

Take the constant out:

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

= \frac{1}{4} \frac{d}{d x} (x^{4} + 8)

Apply the Sum/Difference Rule:

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

= \frac{1}{4} (\frac{d}{d x} (x^{4}) + \frac{d}{d x} (8))

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

= \frac{1}{4} (4 x^{3} + 0)

Simplify

\frac{1}{4} (4 x^{3} + 0) : x^{3}

= x^{3}

g (x) = 3 x^{4} - 140 x^{3}

\int \frac{1}{t} d t

\frac{d}{d x} (4 x^{3} y^{- 2})

\frac{d}{d x} (cos^{n} (x))

\frac{\partial}{\partial x} (- x)

What is the derivative of 1/4 (x^4+8) ?
The derivative of 1/4 (x^4+8) is x^3
What is the first derivative of 1/4 (x^4+8) ?
The first derivative of 1/4 (x^4+8) is x^3