What is the derivative of f(x)=(1+8x^2)(x-x^2) ?

The derivative of f(x)=(1+8x^2)(x-x^2) is -32x^3+24x^2-2x+1

What is the first derivative of f(x)=(1+8x^2)(x-x^2) ?

The first derivative of f(x)=(1+8x^2)(x-x^2) is -32x^3+24x^2-2x+1

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derivative of

f (x) = (1 + 8 x^{2}) (x - x^{2})

- 32 x^{3} + 24 x^{2} - 2 x + 1

Solution steps

\frac{d}{d x} ((1 + 8 x^{2}) (x - x^{2}))

Apply the Product Rule:

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

f = 1 + 8 x^{2}, g = x - x^{2}

= \frac{d}{d x} (1 + 8 x^{2}) (x - x^{2}) + \frac{d}{d x} (x - x^{2}) (1 + 8 x^{2})

\frac{d}{d x} (1 + 8 x^{2}) = 16 x

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

= 16 x (x - x^{2}) + (1 - 2 x) (1 + 8 x^{2})

Simplify

16 x (x - x^{2}) + (1 - 2 x) (1 + 8 x^{2}) : - 32 x^{3} + 24 x^{2} - 2 x + 1

= - 32 x^{3} + 24 x^{2} - 2 x + 1

\frac{d}{d x} (e^{x - 8})

\frac{d}{d x} (a^{x} + x^{n})

\frac{\partial}{\partial x} (6 x e^{x y})

\frac{\partial}{\partial x} (15 x^{2} + y^{2})

\int 2 y (- y + 2) d y

What is the derivative of f(x)=(1+8x^2)(x-x^2) ?
The derivative of f(x)=(1+8x^2)(x-x^2) is -32x^3+24x^2-2x+1
What is the first derivative of f(x)=(1+8x^2)(x-x^2) ?
The first derivative of f(x)=(1+8x^2)(x-x^2) is -32x^3+24x^2-2x+1