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

The derivative of-x(2x-y^3) is -4x+3xy^2(dy)/(dx)+y^3

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

The first derivative of-x(2x-y^3) is -4x+3xy^2(dy)/(dx)+y^3

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

- 4 x + 3 x y^{2} \frac{d y}{d x} + y^{3}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

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

f = x, g = 2 x - y^{3}

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

\frac{d x}{d x} = 1

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

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

Simplify

- (1 \cdot (2 x - y^{3}) + (2 - 3 y^{2} \frac{d y}{d x}) x) : - 4 x + 3 x y^{2} \frac{d y}{d x} + y^{3}

= - 4 x + 3 x y^{2} \frac{d y}{d x} + y^{3}

\int (4 + x) d x

e^{\frac{1}{θ}}

n = 0 \sum \infty \frac{3}{n n}

\int \frac{sin ( 2 x )}{1 + cos ^{2} ( 2 x )} d x

y^{^{'}} = y (2 + y) t, y (0) = 1

What is the derivative of-x(2x-y^3) ?
The derivative of-x(2x-y^3) is -4x+3xy^2(dy)/(dx)+y^3
What is the first derivative of-x(2x-y^3) ?
The first derivative of-x(2x-y^3) is -4x+3xy^2(dy)/(dx)+y^3