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

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

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

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

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derivative of

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

4 x^{3} - 6 x^{2} + 2 x - 2

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

ln (1 - x^{2})

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

\int_{0}^{1} \frac{1 + 12 t}{1 + 3 t} d t

(- 6 sin (6 x))^{^{'}}

\frac{d}{d x} (\frac{1}{8 x + 3})

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