What is the derivative of-x^2y^2 ?

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

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

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

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

- 2 x y^{2} - 2 x^{2} y \frac{d y}{d x}

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

= - 2 x y^{2} - 2 x^{2} y \frac{d y}{d x}

x \to 0 lim (x cos (\frac{14}{x}))

x \to 1 - lim (3 x - 2)

\frac{\partial}{\partial u} (u + v)

\int \frac{x ^{2}}{2} \frac{1}{x} d x

t^{2} \cdot sin (3 t)

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