What is the derivative of x^4-8x^2+a ?

The derivative of x^4-8x^2+a is 4x^3-16x

What is the first derivative of x^4-8x^2+a ?

The first derivative of x^4-8x^2+a is 4x^3-16x

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\frac{d}{d x} (x^{4} - 8 x^{2} + a)

4 x^{3} - 16 x

Solution steps

\frac{d}{d x} (x^{4} - 8 x^{2} + a)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) - \frac{d}{d x} (8 x^{2}) + \frac{d a}{d x}

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

\frac{d}{d x} (8 x^{2}) = 16 x

\frac{d a}{d x} = 0

= 4 x^{3} - 16 x + 0

Simplify

= 4 x^{3} - 16 x

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

x \to - 19 + lim (k (x))

4 sin (π \cdot x)

\int_{0}^{1} x^{- 2} d x

What is the derivative of x^4-8x^2+a ?
The derivative of x^4-8x^2+a is 4x^3-16x
What is the first derivative of x^4-8x^2+a ?
The first derivative of x^4-8x^2+a is 4x^3-16x