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

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

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

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

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derivative of

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

3 x^{2} + 2 x - 4

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (x^{2} - 4) = 2 x

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

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

Simplify

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

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

4 x, x^{2} - 12

t^{2} y^{^{''}} + 7 t y^{^{'}} + 5 y = 0

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

\int_{1}^{4} \frac{1}{x ^{2}} d x

\frac{d}{d x} (sin (x) - \frac{1}{18} sin (3 x))

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