What is the derivative of y=-x^3(3x^4-2) ?

The derivative of y=-x^3(3x^4-2) is -21x^6+6x^2

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

The first derivative of y=-x^3(3x^4-2) is -21x^6+6x^2

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derivative of

y = - x^{3} (3 x^{4} - 2)

- 21 x^{6} + 6 x^{2}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3}, g = 3 x^{4} - 2

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

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

\frac{d}{d x} (3 x^{4} - 2) = 12 x^{3}

= - (3 x^{2} (3 x^{4} - 2) + 12 x^{3} x^{3})

Simplify

- (3 x^{2} (3 x^{4} - 2) + 12 x^{3} x^{3}) : - 21 x^{6} + 6 x^{2}

= - 21 x^{6} + 6 x^{2}

What is the derivative of y=-x^3(3x^4-2) ?
The derivative of y=-x^3(3x^4-2) is -21x^6+6x^2
What is the first derivative of y=-x^3(3x^4-2) ?
The first derivative of y=-x^3(3x^4-2) is -21x^6+6x^2