What is the derivative of y=(x+5)(x^3-1)(x^2-x+3) ?

The derivative of y=(x+5)(x^3-1)(x^2-x+3) is 6x^5+20x^4-8x^3+42x^2-8x+2

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

The first derivative of y=(x+5)(x^3-1)(x^2-x+3) is 6x^5+20x^4-8x^3+42x^2-8x+2

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derivative of

y = (x + 5) (x^{3} - 1) (x^{2} - x + 3)

6 x^{5} + 20 x^{4} - 8 x^{3} + 42 x^{2} - 8 x + 2

Solution steps

\frac{d}{d x} ((x + 5) (x^{3} - 1) (x^{2} - x + 3))

Apply the Product Rule:

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

f = x + 5, g = (x^{3} - 1) (x^{2} - x + 3)

= \frac{d}{d x} (x + 5) (x^{3} - 1) (x^{2} - x + 3) + \frac{d}{d x} ((x^{3} - 1) (x^{2} - x + 3)) (x + 5)

\frac{d}{d x} (x + 5) = 1

\frac{d}{d x} ((x^{3} - 1) (x^{2} - x + 3)) = 5 x^{4} - 4 x^{3} + 9 x^{2} - 2 x + 1

= 1 \cdot (x^{3} - 1) (x^{2} - x + 3) + (5 x^{4} - 4 x^{3} + 9 x^{2} - 2 x + 1) (x + 5)

Simplify

1 \cdot (x^{3} - 1) (x^{2} - x + 3) + (5 x^{4} - 4 x^{3} + 9 x^{2} - 2 x + 1) (x + 5) : 6 x^{5} + 20 x^{4} - 8 x^{3} + 42 x^{2} - 8 x + 2

= 6 x^{5} + 20 x^{4} - 8 x^{3} + 42 x^{2} - 8 x + 2

\frac{d}{d x} (2 e^{7 x})

\int \frac{1}{5 - 4 sin ( x ) + 3 cos ( x )} d x

- 4 y^{^{'}} + y = 1

e^{x}, - x + 1, x = 1

\frac{d}{d x} ((2 x + 5)^{2})

What is the derivative of y=(x+5)(x^3-1)(x^2-x+3) ?
The derivative of y=(x+5)(x^3-1)(x^2-x+3) is 6x^5+20x^4-8x^3+42x^2-8x+2
What is the first derivative of y=(x+5)(x^3-1)(x^2-x+3) ?
The first derivative of y=(x+5)(x^3-1)(x^2-x+3) is 6x^5+20x^4-8x^3+42x^2-8x+2