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

The derivative of (1+4x^2)(x-x^2) is -16x^3+12x^2-2x+1

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

The first derivative of (1+4x^2)(x-x^2) is -16x^3+12x^2-2x+1

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derivative of

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

- 16 x^{3} + 12 x^{2} - 2 x + 1

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (1 + 4 x^{2}) = 8 x

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

= 8 x (x - x^{2}) + (1 - 2 x) (1 + 4 x^{2})

Simplify

8 x (x - x^{2}) + (1 - 2 x) (1 + 4 x^{2}) : - 16 x^{3} + 12 x^{2} - 2 x + 1

= - 16 x^{3} + 12 x^{2} - 2 x + 1

\frac{d}{d x} (A cos (x) + B sin (x))

\frac{d}{d x} (- \frac{2}{9 x ^{\frac{4}{3}}})

\int_{- 1}^{6} - x^{2} + 5 x + 6 d x

x \to 0 + lim (\frac{∣ 8 x ∣}{x})

y = sin^{2} (π x), [0, 1]

What is the derivative of (1+4x^2)(x-x^2) ?
The derivative of (1+4x^2)(x-x^2) is -16x^3+12x^2-2x+1
What is the first derivative of (1+4x^2)(x-x^2) ?
The first derivative of (1+4x^2)(x-x^2) is -16x^3+12x^2-2x+1