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

The derivative of y=(6-x^2)(x^3-3x+4) is -5x^4+27x^2-8x-18

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

The first derivative of y=(6-x^2)(x^3-3x+4) is -5x^4+27x^2-8x-18

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derivative of

y = (6 - x^{2}) (x^{3} - 3 x + 4)

- 5 x^{4} + 27 x^{2} - 8 x - 18

Solution steps

\frac{d}{d x} ((6 - x^{2}) (x^{3} - 3 x + 4))

Apply the Product Rule:

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

f = 6 - x^{2}, g = x^{3} - 3 x + 4

= \frac{d}{d x} (6 - x^{2}) (x^{3} - 3 x + 4) + \frac{d}{d x} (x^{3} - 3 x + 4) (6 - x^{2})

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

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

= (- 2 x) (x^{3} - 3 x + 4) + (3 x^{2} - 3) (6 - x^{2})

Simplify

(- 2 x) (x^{3} - 3 x + 4) + (3 x^{2} - 3) (6 - x^{2}) : - 5 x^{4} + 27 x^{2} - 8 x - 18

= - 5 x^{4} + 27 x^{2} - 8 x - 18

\int ∣ u + 1 ∣ d u

\frac{\partial}{\partial t} (r sin (t) sin (x))

\frac{d}{d x} (e^{2 x - y})

\frac{d}{d y} (10 - 6 y + e^{- 3 x})

\int e^{- \frac{x}{20}} d x

What is the derivative of y=(6-x^2)(x^3-3x+4) ?
The derivative of y=(6-x^2)(x^3-3x+4) is -5x^4+27x^2-8x-18
What is the first derivative of y=(6-x^2)(x^3-3x+4) ?
The first derivative of y=(6-x^2)(x^3-3x+4) is -5x^4+27x^2-8x-18