What is the derivative of (2x^2+2)^3(x^2-1)^2 ?

The derivative of (2x^2+2)^3(x^2-1)^2 is 12x(2x^2+2)^2(x^2-1)^2+4x(x^2-1)(2x^2+2)^3

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

The first derivative of (2x^2+2)^3(x^2-1)^2 is 12x(2x^2+2)^2(x^2-1)^2+4x(x^2-1)(2x^2+2)^3

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derivative of

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

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

Solution steps

\frac{d}{d x} ((2 x^{2} + 2)^{3} (x^{2} - 1)^{2})

Apply the Product Rule:

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

f = (2 x^{2} + 2)^{3}, g = (x^{2} - 1)^{2}

= \frac{d}{d x} ((2 x^{2} + 2)^{3}) (x^{2} - 1)^{2} + \frac{d}{d x} ((x^{2} - 1)^{2}) (2 x^{2} + 2)^{3}

\frac{d}{d x} ((2 x^{2} + 2)^{3}) = 12 x (2 x^{2} + 2)^{2}

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

= 12 x (2 x^{2} + 2)^{2} (x^{2} - 1)^{2} + 4 x (x^{2} - 1) (2 x^{2} + 2)^{3}

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

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 7

\frac{\partial}{\partial x} (x^{2} - y^{3} - x^{2} y + y)

\frac{\partial}{\partial x} (- 2 (1 - x) - 40 x (y - x^{2}))

\frac{d}{d x} (e (- x))

What is the derivative of (2x^2+2)^3(x^2-1)^2 ?
The derivative of (2x^2+2)^3(x^2-1)^2 is 12x(2x^2+2)^2(x^2-1)^2+4x(x^2-1)(2x^2+2)^3
What is the first derivative of (2x^2+2)^3(x^2-1)^2 ?
The first derivative of (2x^2+2)^3(x^2-1)^2 is 12x(2x^2+2)^2(x^2-1)^2+4x(x^2-1)(2x^2+2)^3