What is the derivative of f(x)=4x^4+3x^2-1 ?

The derivative of f(x)=4x^4+3x^2-1 is 16x^3+6x

What is the first derivative of f(x)=4x^4+3x^2-1 ?

The first derivative of f(x)=4x^4+3x^2-1 is 16x^3+6x

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derivative of

f (x) = 4 x^{4} + 3 x^{2} - 1

16 x^{3} + 6 x

Solution steps

\frac{d}{d x} (4 x^{4} + 3 x^{2} - 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{4}) + \frac{d}{d x} (3 x^{2}) - \frac{d}{d x} (1)

\frac{d}{d x} (4 x^{4}) = 16 x^{3}

\frac{d}{d x} (3 x^{2}) = 6 x

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

= 16 x^{3} + 6 x - 0

Simplify

= 16 x^{3} + 6 x

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What is the derivative of f(x)=4x^4+3x^2-1 ?
The derivative of f(x)=4x^4+3x^2-1 is 16x^3+6x
What is the first derivative of f(x)=4x^4+3x^2-1 ?
The first derivative of f(x)=4x^4+3x^2-1 is 16x^3+6x