What is the derivative of x^4-8x^2+1 ?

The derivative of x^4-8x^2+1 is 4x^3-16x

What is the first derivative of x^4-8x^2+1 ?

The first derivative of x^4-8x^2+1 is 4x^3-16x

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\frac{d}{d x} (x^{4} - 8 x^{2} + 1)

4 x^{3} - 16 x

Solution steps

\frac{d}{d x} (x^{4} - 8 x^{2} + 1)

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}{d x} (1)

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

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

\frac{d}{d x} (1) = 0

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

Simplify

= 4 x^{3} - 16 x

\int arcsin (y) d y

\int_{- \infty}^{0} \frac{1}{1 + x ^{2}} d x

\frac{d}{d x} (ln (5 x + 1))

f (x) = \frac{19}{x ^{5}}

\frac{d}{d x} (- 2 lo g_{6} (x))

What is the derivative of x^4-8x^2+1 ?
The derivative of x^4-8x^2+1 is 4x^3-16x
What is the first derivative of x^4-8x^2+1 ?
The first derivative of x^4-8x^2+1 is 4x^3-16x