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

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

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

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

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

x y (x \frac{d y}{d x} + y)

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \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{1}{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}

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

Simplify

\frac{1}{2} (2 x y^{2} + 2 y \frac{d y}{d x} x^{2}) : x y (x \frac{d y}{d x} + y)

= x y (x \frac{d y}{d x} + y)

\int - 4 x^{6} - 5 x^{5} - 2 x^{2} + 5 - e^{x} d x

f (x) = arcsec (x + 2)

x 6^{x}

\frac{\partial}{\partial x} (\frac{1}{x + 3})

x \to 0 lim (x^{2} - 1)

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