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

The derivative of y=(3-x^2)(x^3-2x+1) is -5x^4+15x^2-2x-6

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

The first derivative of y=(3-x^2)(x^3-2x+1) is -5x^4+15x^2-2x-6

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derivative of

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

- 5 x^{4} + 15 x^{2} - 2 x - 6

Solution steps

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

Apply the Product Rule:

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

f = 3 - x^{2}, g = x^{3} - 2 x + 1

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

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

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

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

Simplify

(- 2 x) (x^{3} - 2 x + 1) + (3 x^{2} - 2) (3 - x^{2}) : - 5 x^{4} + 15 x^{2} - 2 x - 6

= - 5 x^{4} + 15 x^{2} - 2 x - 6

\frac{d x}{d t} = \frac{1 - x}{k}

f (x) = sin (7 ln (x))

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{2}) = 0

n = 0 \sum \infty (x + 3)^{n}

\int_{0}^{\frac{π}{2}} 6 sin^{2} (3 x) d x

What is the derivative of y=(3-x^2)(x^3-2x+1) ?
The derivative of y=(3-x^2)(x^3-2x+1) is -5x^4+15x^2-2x-6
What is the first derivative of y=(3-x^2)(x^3-2x+1) ?
The first derivative of y=(3-x^2)(x^3-2x+1) is -5x^4+15x^2-2x-6