What is the derivative of x^4+8 ?

The derivative of x^4+8 is 4x^3

What is the first derivative of x^4+8 ?

The first derivative of x^4+8 is 4x^3

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

4 x^{3}

Solution steps

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

Apply the Sum/Difference Rule:

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

= \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

= 4 x^{3} + 0

Simplify

= 4 x^{3}

x \to - \infty lim (5 + x^{2})

\frac{d y}{d x} = x y

y^{^{''}} + 4 y + 8 = 0, y (0) = 4, y^{^{'}} (0) = - 3

f (x) = - 2 \cdot x, at x = 1

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