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

The derivative of y=(4x^2+x)(x-x^2) is -16x^3+9x^2+2x

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

The first derivative of y=(4x^2+x)(x-x^2) is -16x^3+9x^2+2x

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derivative of

y = (4 x^{2} + x) (x - x^{2})

- 16 x^{3} + 9 x^{2} + 2 x

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

(8 x + 1) (x - x^{2}) + (1 - 2 x) (4 x^{2} + x) : - 16 x^{3} + 9 x^{2} + 2 x

= - 16 x^{3} + 9 x^{2} + 2 x

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

x \to 3 lim (+ (- 1))

\frac{d}{d t} (e^{7 t} e^{4 i t})

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

\frac{\partial}{\partial x} (arctan (x y^{2}))

What is the derivative of y=(4x^2+x)(x-x^2) ?
The derivative of y=(4x^2+x)(x-x^2) is -16x^3+9x^2+2x
What is the first derivative of y=(4x^2+x)(x-x^2) ?
The first derivative of y=(4x^2+x)(x-x^2) is -16x^3+9x^2+2x