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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

ln (1 + x^{4})

\frac{d}{d x} ((lo g_{10} (6) \cdot x^{2})^{3})

(3 x^{2} - 5 x)^{45}

cos (2 x + 3)

(\frac{x}{3})^{2}

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