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

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

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

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

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

4 x^{3} + 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 4 x^{3} + 8

f (x) = - \frac{4}{x}

x^{3}, at x = 1

y = (x^{6} - 10) x

\frac{d y}{d t} = 0.0572 (400 - y), y (0) = 32

y^{^{'}} + \frac{1 - 2 x}{x ^{2}} y - 1 = 0

What is the derivative of x^4+8x ?
The derivative of x^4+8x is 4x^3+8
What is the first derivative of x^4+8x ?
The first derivative of x^4+8x is 4x^3+8