What is (\partial)/(\partial x)((x^4-1)(x^2+1)) ?

The answer to (\partial)/(\partial x)((x^4-1)(x^2+1)) is 6x^5+4x^3-2x

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(\partial)/(\partial x)((x^4-1)(x^2+1))

\frac{\partial}{\partial x} ((x^{4} - 1) (x^{2} + 1))

6 x^{5} + 4 x^{3} - 2 x

Solution steps

\frac{\partial}{\partial x} ((x^{4} - 1) (x^{2} + 1))

Apply the Product Rule:

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

f = x^{4} - 1, g = x^{2} + 1

= \frac{\partial}{\partial x} (x^{4} - 1) (x^{2} + 1) + \frac{\partial}{\partial x} (x^{2} + 1) (x^{4} - 1)

\frac{\partial}{\partial x} (x^{4} - 1) = 4 x^{3}

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

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

Simplify

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

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

\frac{\partial}{\partial y} (y sin (x))

x \to 2 + lim (2^{x} - 4)

x \to 1 + lim (ln^{x} (x))

\int \frac{4 \cdot x + 3}{x ^{2} - 25} d x

y (x + 3) + \frac{d y}{d x} = 0

What is (\partial)/(\partial x)((x^4-1)(x^2+1)) ?
The answer to (\partial)/(\partial x)((x^4-1)(x^2+1)) is 6x^5+4x^3-2x