What is the derivative of x^{1/2}y^{1/2} ?

The derivative of x^{1/2}y^{1/2} is (y^{1/2})/(2x^{1/2)}+(x^{1/2}\frac{dy)/(dx)}{2y^{1/2}}

What is the first derivative of x^{1/2}y^{1/2} ?

The first derivative of x^{1/2}y^{1/2} is (y^{1/2})/(2x^{1/2)}+(x^{1/2}\frac{dy)/(dx)}{2y^{1/2}}

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

\frac{y ^{\frac{1}{2}}}{2 x ^{\frac{1}{2}}} + \frac{x ^{\frac{1}{2}} \frac{d y}{d x}}{2 y ^{\frac{1}{2}}}

Solution steps

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

Apply the Product Rule:

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

f = x^{\frac{1}{2}}, g = y^{\frac{1}{2}}

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

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

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

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

Simplify

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

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

x \to 8 lim (x^{3} - 6 x + 5)

\frac{1}{x ^{2}}, at x = 3

\frac{d y}{d x} 3 x^{3} + 3 y^{3} - 5 x y = 0

4 x^{6} e^{x}

\int \frac{t}{a t ^{2} + b} d t

What is the derivative of x^{1/2}y^{1/2} ?
The derivative of x^{1/2}y^{1/2} is (y^{1/2})/(2x^{1/2)}+(x^{1/2}\frac{dy)/(dx)}{2y^{1/2}}
What is the first derivative of x^{1/2}y^{1/2} ?
The first derivative of x^{1/2}y^{1/2} is (y^{1/2})/(2x^{1/2)}+(x^{1/2}\frac{dy)/(dx)}{2y^{1/2}}