What is the derivative of x^4-8x^2+3 ?

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

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

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

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\frac{d}{d x} (x^{4} - 8 x^{2} + 3)

4 x^{3} - 16 x

Solution steps

\frac{d}{d x} (x^{4} - 8 x^{2} + 3)

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} (3)

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

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

\frac{d}{d x} (3) = 0

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

Simplify

= 4 x^{3} - 16 x

x^{3} y^{^{'}} + x^{2} y = x^{7} y^{\frac{3}{4}}

\frac{d}{d x} (- 7 e^{- 7 x})

x \to 4 lim (ln (x + 2))

\int \frac{x}{( x + 1 ) ^{2} ( x + 2 ) ^{2}} d x

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