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

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

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

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

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derivative of

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

4 x^{3}

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

= 4 x^{3}

n = 1 \sum \infty \frac{x ^{n}}{1 1 ^{n}}

\int 2 (x - 2) d x

\int cos (t - x) d x

f (x) = 7 x - 5

\int (2 x + 4)^{2} d x

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