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

The derivative of 4x^4+3x^2 is 16x^3+6x

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

The first derivative of 4x^4+3x^2 is 16x^3+6x

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

16 x^{3} + 6 x

Solution steps

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

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} (4 x^{4}) = 16 x^{3}

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

= 16 x^{3} + 6 x

f (x) = 5 + x^{2}

\int_{0}^{\frac{1}{2}} t^{2} d t

\frac{\partial}{\partial x} (8 + e^{- x y} cos (y))

f (x) = \frac{1}{2 + 3 x}, (2, \frac{1}{8})

y = 1 - 2 x

What is the derivative of 4x^4+3x^2 ?
The derivative of 4x^4+3x^2 is 16x^3+6x
What is the first derivative of 4x^4+3x^2 ?
The first derivative of 4x^4+3x^2 is 16x^3+6x